Polygon Ratio Problem: Many Proofs, One Degree of Freedom¶

Author: Changchun Shi
Date: First version Jan 12, 2026
Last update Jan 25, 2026

Introduction¶

During the Christmas holidays, a basic geometry proof problem showed up in my class alumni chat (Figure 1 is the original screenshot from classmate Lin He). He asked about the second question because the first is almost trivial, and that set the discussion in motion. I have always liked Euclidean geometry and excelled in my teenage competitions (I don't recall failing to prove any geometry problems in that period). Problems like this rarely stop at one proof. Draw one line and a ratio drops out; draw another and a different argument appears. What follows is a cleaned collection: one quick sine-law proof, several synthetic proofs, an analytic-geometry proof (with help from ChatGPT 5.2-thinking, edited for clarity), and a one-parameter generalization that reveals deeper structure. I hope it gives younger students a sense of how much structure a single diagram can hide. I also plan to use this pilot problem to introduce state-of-the-art AI approaches for solving such problems in separate notes.

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Figure 1: The original problem (photo from the chat).

Let me first restate the problem in words and redraw the diagram more cleanly.

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Figure 2: A clean redraw in GeoGebra.

Problem statement (translated). In trapezoid $ABCD$, $AD\parallel BC$, $ngle C=90^\circ$, and $BC=CD$. Point $E$ lies on $CD$, and $DE=AD$. Through $E$, draw $EF\perp AB$ with $F$ as the foot of the perpendicular. Draw $BD$; it intersects $AE$ and $EF$ at points $G$ and $H$, respectively. Prove:

  1. $BF\cdot HE=GE\cdot BH$.
  2. $BE\cdot DH=BD\cdot EH$.

Part 1. Synthetic geometry (auxiliary lines)¶

The basic equal angles and lengths are shown in Figure 3.

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Figure 3: Basic angle and length relations.

Question (1) is a warm-up; it follows from similar triangles visible in Figure 3. For part (2), a quick start is the sine law. For a triangle $ABC$ with side lengths $a,b,c$ opposite vertices $A,B,C$, we have $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.$$ This gives Proof 0 immediately.

Proof 0 (sine law)¶

The ratios $DB:BE$ and $DH:HE$ are both $\sin(45^\circ+\alpha):\sin 45^\circ$, so they are equal. This proof uses no auxiliary lines.

To avoid assuming the sine law, I drew auxiliary lines. Rather than separate diagrams per proof, the next figure collects all extra points. Each added point carries a subscript that matches the proof number (for example, $D_1$ is used only in Proof 1).

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Figure 4: All auxiliary lines used in the synthetic proofs.

Below are brief proof sketches. Each can be completed as a standard middle-school geometry exercise.

Proof 1¶

Extend $DC$ to $D_1$ so that $D_1$ is symmetric to $D$ with respect to $BC$. Then $\triangle EBD_1\sim\triangle EHD$, so the desired ratio follows.

Proof 2¶

Extend $DC$ to $E_2$ so that $E_2$ is symmetric to $E$ with respect to $BC$. Then $\triangle E_2BD\sim\triangle EHD$, so the desired ratio follows.

Proof 3¶

From $H$ draw a line parallel to $BE$ and intersect $DE$ at $E'$. Then $HE=HE'$. The rest follows.

Proof 4¶

Extend $AE$ to meet $BC$ at $C_4$. Then $AC_4=BD$. Also $\triangle HED\sim\triangle ABC_4$, so the ratio follows. (Credit: Prof. Xueheng Lan.)

Proof 4.1¶

Use the reflection property: the ray from $B$ to $E$ reflects across $DE$ and passes through $H$. This gives the same angle relation at $DE$, so the ratio follows.

Proof 5¶

Draw a line from $H$ parallel to $DE$ and intersect $EA$ at $H_5$. Then $DH=EH_5$, and $\triangle EHH_5\sim\triangle BED$, which leads to the result.

Proof 6¶

Draw a line from $H$ parallel to $BC$ and intersect $DE$ at $H_6$. Then $DH/DB=HH_6/BD=EH/BD$.

Proof 7¶

Draw a line from $H$ parallel to $DE$ and intersect $BE$ at $H_7$. Then $HE:BE=H_7E:BE=HD:BD$.

Proof 8¶

The distances from $D$ to $BE$ and $EH$ are equal (use $DD_8$ and $DD'_8$; these are non-essential since $\angle BEC=\angle HED=45^\circ+\alpha$). The distances from $E$ to $BD$ and $HD$ are also equal. So area $BDE$ to area $DHE$ can be expressed as $BE\cdot DD_8/2:EH\cdot DD'_8/2=BE:EH$, and also as $BD\cdot GE/2:DH\cdot GE/2=BD:DH$. Hence they are equal.

Proof 9¶

Extend $BE$ to $A_9$ where $EA_9=EH$. Then $DH=DA_9$, $EH=EA_9$, and $\angle EDA_9=\angle EDH=45^\circ$. Since $DE$ bisects $\angle BDA_9$, we get $BD:DA_9=BE:EA_9$. The rest is trivial.

Proof 10¶

Extend $EH$ to $B_{10}$ where $B_{10}$ intersects the extension of line $BC$. Then $B_{10}$ is the reflection of $B$ across $BC$ (since $HE$ is the reflection path of light $BE$ at mirror $DC$). One can show $\angle CDB_{10}=\angle BDC=45^\circ$, so $DH:HE=DB_{10}:EB_{10}=DB:BE$.

Proof 11¶

This one also needs no auxiliary lines: notice $HD=HG+GE$ and $BD=BG+GE$, and $\triangle HGE\sim\triangle EGB$, so the result follows.

A note on degrees of freedom: the only adjustable angle in the original configuration is $\alpha<45^\circ$. If $\alpha=45^\circ$, then $H$ and $B$ overlap and the result is trivial. If $\alpha>45^\circ$, then $H$ lies on the extension of $DB$, and $E$ lies on the extension of $DC$. The roles of $B$ and $H$ swap, and the same style of arguments still works. Similarly, one can show that a ray from $B$ to $E$ reflects across $DE$ and reaches $H$.

Other proofs¶

By now we should be done. However, if you followed the solutions so far, you likely noticed two patterns: (1) we have essentially drawn a square $BDB_{10}D_1$ in Figure 4, and many proofs are based on various points on this square; (2) polygon $ABED$ is symmetric about $BD$, which is why we could leverage $AB=BE$ and $AH=HE$ in several steps. It follows that any proof obtained by extending lines in the original diagram has a mirror proof across $BD$. These are indicated again by subscripts, and they give Proofs 12, 13, 14, 15, and 16. They form the symmetric rectangle $BDB_{15}D_{12}$ and are shown in Figure 5.

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Figure 5: Five additional symmetric proofs.

Finally, several proofs such as Proofs 3, 5, 6, and 7 focus on what happens inside polygon $ABED$. Each has a symmetric counterpart (Proofs 17, 18, 19, 20) reflected across $BD$. For clarity, these are not displayed.

Part 2. Analytic geometry proof¶

This is a short coordinate proof.

Goal¶

$$BE\cdot DH = BD\cdot EH.$$

Proof¶

Setup. Let $C=(0,0)$, $B=(-1,0)$, $D=(0,1)$. Then $AD\parallel BC$ implies $A=(t-1,1)$ when $E=(0,t)$ and $DE=AD$ with $0<t<1$.

Lines. $BD: y=x+1$. The slope of $AB$ is $1/t$, so $EF\perp AB$ has slope $-t$ and passes through $E$: $EF: y=t-tx$.

Intersection. Solve $x+1=t-tx$ to get $$H=(\frac{t-1}{t+1},\ \frac{2t}{t+1}).$$

Common factor. $$H-D=\frac{t-1}{t+1}(1,1),\qquad H-E=\frac{t-1}{t+1}(1,-t).$$ Hence $$DH=\frac{1-t}{t+1}\sqrt{2},\qquad EH=\frac{1-t}{t+1}\sqrt{1+t^2}.$$ Also $BD=\sqrt{2}$ and $BE=\sqrt{1+t^2}$.

Finish. $$BE\cdot DH=\sqrt{1+t^2}\cdot \frac{1-t}{t+1}\sqrt{2}=BD\cdot EH.$$

Conclusion: $BE\cdot DH = BD\cdot EH$.

Part 3. One more degree of freedom (ellipse view)¶

The $45^\circ$ angle is special. If we relax that angle, the clean right-angle structure disappears, but the problem still has a simple geometric core: reflection on an ellipse. The key insight is to generalize the reflection: the ray from $B$ to $E$ reflects across $DE$ and heads to $H$. This reveals an ellipse that reflects light from one focus to the other at any point $E$ on it. That reflection property is equivalent to the constant-sum-of-distances definition of an ellipse.

Construction¶

Let $O$ be the ellipse with foci $B$ and $H$. Take a point $A$ on $O$, with $A$ closer to $H$. Draw the tangent line $AD$ at $A$, and let it intersect $BH$ at $D$. Take a point $E$ on $O$, the reflection of $A$ with respect to the major axis. Through $E$ draw the line $D-E-C$, and choose $C$ so that $BC\parallel AD$. Draw $EH$, and let it intersect $AB$ at $F$. In general, $\angle BFE$ is not $90^\circ$, but when $\theta=45^\circ$, $\angle BFE=90^\circ$. The resulting shape is shown in Figure 6.

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Figure 6: Ellipse construction with foci $B$ and $H$.

Proof 3.1 Proving the equal ratio using areflection property of ellipse¶

By the reflection property of the ellipse (a light ray from $B$ to $E$ reflects and then passes through $H$), we obtain $$\frac{EH}{HD}=\frac{BE}{BD}.$$

Reflect $H$ across line $DE$ to $H_m$. Then $$\frac{DB}{BE}=\frac{DH_m}{EH_m}=\frac{DH}{EH},$$ which is the same proportion.

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Figure 7: Reflection proof for the ellipse construction.

In this generalized setup, $EF$ is no longer perpendicular to $AB$. In fact, the perpendicular line through $E$ intersects $AB$ and $DB$ at $F'$ and $H'$, as shown in Figure 7. Here $H'$ differs from $H$ and does not satisfy the ratio property. The original problem is special because the $45^\circ$ angles create a surplus of right angles. The single degree of freedom here is essentially the ellipse eccentricity, with $A$ and $C$ determined by the right angle $\angle ADC$. When the ellipse degenerates to a circle (foci coincide), the identity becomes obvious.

In this setup, points $A$, $F$, $C$, and $G$, and the lines through them, are not even essential; only the foci $B$ and $H$, and the tangent line $DE$, are necessary, with $D$ on the major axis. This shows the true simplicity of the original problem; Figure 8 shows the pure geometric approach.

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Figure 8: The simplified ellipse setup, which reveals the essence of the original problem.

The diagram also yields a second, similarity-based proof.

Proof 3.2, similarity¶

Let $T$ and $S$ be the feet of the perpendiculars from $H$ and $B$ to the tangent line $DE$. By the reflection property, $\triangle HET\sim\triangle BES$, so $HE:BE=HT:BS$. Likewise, $\triangle HDT\sim\triangle BDS$, so $HD:DB=HT:BS$. Combining gives $HD/DB=HE/BE$.

Interestingly, we can also compute the ratio analytically in the ellipse model. It turns out to be simplest using the de La Hire (eccentric-angle) parameter, revealing the geometry transparently.

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Figure 9: de La Hire parameters and outer-circle view with $E$, $E'$, and $E''$.

Proof 3.3 (de La Hire + outer circle)¶

Let the ellipse be $x^{2}/a^{2}+y^{2}/b^{2}=1$ with center $O$, foci $B=(-c,0)$ and $H=(c,0)$, and $c^{2}=a^{2}-b^{2}$. Use the de La Hire (eccentric-angle) parameter $\beta$: $$ E=(a\cos\beta,\, b\sin\beta),\qquad EB=a+c\cos\beta,\qquad EH=a-c\cos\beta $$ (up to swapping $B,H$). Let $\ell$ be the tangent at $E$ and $D=\ell\cap BH$.

From the tangent formula $$ \frac{xx_0}{a^{2}}+\frac{yy_0}{b^{2}}=1 \quad\Rightarrow\quad \frac{x\cos\beta}{a}+\frac{y\sin\beta}{b}=1, $$ setting $y=0$ gives $x_D=a/\cos\beta$.

Geometrically, the affine map $(x,y)\mapsto(x,\frac{a}{b}y)$ sends the ellipse to the outer circle $x^{2}+y^{2}=a^{2}$ and sends $E$ to $E'=(a\cos\beta,a\sin\beta)$. Since only $y$ is scaled, $E$ and $E'$ lie on a line perpendicular to $BH$ and share the same $x$-coordinate. Let $E''$ be the intersection of the line through $E$ parallel to $BH$ with the outer circle, so $E$ and $E''$ share the same $y$-coordinate (Figure 9). The map preserves tangency, so $\angle{OE'D}$ is right and $OD=a/\cos\beta$.

Therefore $$ DB=\frac{a}{\cos\beta}+c,\qquad DH=\frac{a}{\cos\beta}-c, $$ and $$ \frac{DH}{DB} =\frac{a-c\cos\beta}{a+c\cos\beta} =\frac{EH}{EB}. $$

Corollary 1 (axis-focus scaling; inner structure of the ratio)¶

From $DH=\frac{a}{\cos\beta}-c=\frac{a-c\cos\beta}{\cos\beta}$ and $EH=a-c\cos\beta$, we get $$ \boxed{EH=DH\cos\beta},\qquad \boxed{EB=DB\cos\beta}. $$ This shows that the tangent-axis lengths $DB,DH$ are just the focal lengths $EB,EH$ scaled by the same factor $1/\cos\beta$ (see Figure 9). The main identity $DH/DB=EH/EB$ is exactly the cancellation of that common scale.

Corollary 2 (focus-to-outer-tangent distances)¶

Corollary 1 gives $EH=DH\cos\beta$, and geometrically $DH\cos\beta$ is exactly the perpendicular component of $DH$ to $\ell'$, i.e., $\operatorname{dist}(H,\ell')$ (see Figure 9). Hence $$ \boxed{\operatorname{dist}(H,\ell')=EH},\qquad \boxed{\operatorname{dist}(B,\ell')=EB}. $$

The simple consistency check $$ \operatorname{dist}(B,\ell')+\operatorname{dist}(H,\ell')=2a=2\,\operatorname{dist}(O,\ell') $$ accounts only for the sum. The real content is that each focus-to-tangent distance equals its matching focus-to-point distance; this rigidity is encoded by the de La Hire parameter and the ellipse's affine relation to the circle.

Proof 3.4 (outer circle proof)¶

Once Corollary 2 is in hand, let $P$ and $P'$ be the feet of the perpendiculars from $H$ and $B$ to $\ell'$ (for simplicity Figure 9 only shows $P$). Then $\triangle DHP \sim \triangle DBP'$ (right angles at $P,P'$ and a common angle at $D$), so $DH/DB=HP/BP'=\operatorname{dist}(H,\ell')/\operatorname{dist}(B,\ell')$, recovering the same ratio.

Finally, there is a fun fact. Note that the feet of the perpendiculars drawn in Proof 3.2, $S$ and $T$, are exactly the intersection points of the tangent line $DE$ with the ellipse's outer circle. A quick proof is to show that $OT=a$ (Figure 10). Let $H'$ be the reflection of $H$ across $DE$; then $T$ is the midpoint of $HH'$. Since $O$ is the midpoint of $BH$, we have $OT=\tfrac{1}{2}BH'$. But $BH'=BE+EH=2a$, hence $OT=a$. The same reflection with $B$ gives $OS=a$, so the fun fact is that $S$ and $T$ are precisely the two intersections of $DE$ with the outer circle.

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Figure 10: additional properties of T and S

Also these relations show $OT\parallel BE$ and $HE\parallel OS$. So $HE:OS=DH:DO$ and $OT:BE=DO:DB$. Since $OT=a=OS$, we get $HE:a=DH:DO$ and $a:BE=DO:DB$. So we immediately have two observations, stated below:

Corollary 2 revisit geometrically¶

In Figure 9, $OE'\perp DE'$ and $HP\perp DE'$, so $HP:a=HP:OE'=DH:DO=HE:a$ (the last equality comes from the result above). Therefore $HE=HP=\operatorname{dist}(H,\ell')$; similarly $BE=\operatorname{dist}(B,\ell')$. In this analysis we do not use the parametric expression for $HE$ in terms of $\beta$ and $a,c$. This gives a direct geometric reason for the result stated in Corollary 2.

Furthermore, we get another proof of the distance ratio in this part of the note.

Proof 3.5 (yet another geometric proof)¶

Using the ratio conclusion from our geometric derivation, $HE:a=DH:DO$ and $a:BE=DO:DB$ give us $HE:DH=a:DO$ and $BE:DB=a:DO$ so $HE:BE=DH:DB$. Note that a:DO is simply $\cos\beta$ geometrically since $\angleOE'D$ is right, which fills the geometric explaination of Corollary 1 too and Proof 3.3.

Properties explained in this part of the notes, such as Corollary 2, are not obvious to explain geometrically until one draws extensive auxiliary lines that start to utilize more known geometric property of an ellipse such as the the sum of the distance to foci is $\2a$. This is why the analytical approaches such as proof 3.3 are arguably very effective which heavily leverage algebraic derivations. Through our discussion, a prelude to the rich properties of the ellipse should be apparent emphasizing on using the geometric methods. Historically the ellipse, together with the parabola and the hyperbola, forms the family of conic sections and were studied extensively ever since the ancient greek Apollonius [CONICS-WIKIPEDIA]. Conics found its particular importance in planetary physics since Kepler discovered that planets follow such trajectories. The geometric approaches to the ellipse here echo Isaac Newton's way of proving Kepler's laws from universal gravity proportional to $\hat{\mathbf{r}}/r^2$. For interested readers, see the summary of geometric interpretations of these coordinate-free proofs in [KEPLER-TEACH]. All the geometric properties we concluded in this part can be explained relatively straightforwardly using a parametric representation of the ellipse (whether polar or the traditional Cartesian form), though one coordinate choice can make the representation much easier than another. Again, our two routes to Corollary 2 provide more flavors on the comparisons between analytic approach and the pure geometry approach. The geometric approach reveals a different kind of insight into the subject, often leads to a stronger AHA moment and feels obvious only when the "imaginary" auxiliary lines and circles are discovered.

Part 4. Vector analysis¶

Once one length is fixed, the trapezoid configuration has a single degree of freedom: the angle $\alpha$. In the broader ellipse construction, a second parameter $\theta$ (the tangent point/slope) appears, and the same ratio follows from the reflection property. The special case $\theta=45^\circ$ is exactly the original problem; that is why so many proofs exist, and why each auxiliary line exposes the same invariant ratio in a different way. Part 3 frees the $\theta$ dimension while keeping the reflection property, and that is what ultimately shows the simplicity behind the original picture.

In this part of the note, we focus on other mathematical tools that are useful in geometry. The target ratio is almost trivial here: $DH:DB = EH:EB$, given that the ray $BE$ reflects across the mirror $DE$ and reaches $H$. Part 3 already explained the reflection viewpoint, but it is still instructive to see how these tools express the same invariant.

Since Euclid (~300 BCE), geometry was developed in a synthetic style: one reasons directly with the figure using congruence, similarity, tangency, and auxiliary lines. That is the mindset behind Part 1. A major second language arrived with Descartes (1637): analytic geometry turns geometric constraints into algebra once you choose coordinates, matching the style of Part 2.

This arc also mirrors the historical split between geometry and algebra in physics and mechanics. Newton, in the Principia [NEWTON-PRINCIPIA], favored coordinate-free arguments because they were the clearest language of rigor in 1687, while modern physics often relies on coordinates for universal computation. Analytic geometry existed by Newton's time, but the algebraic language for differential equations matured later; Newton's methods therefore look Euclidean in flavor, mixing auxiliary lines with limiting arguments. Several later authors also gave elementary, coordinate-free proofs (see [KEPLER-TEACH], including Feynman's famous "lost lecture"). It is good for students to appreciate both techniques.

Physics naturally blends geometry with differential ideas. Here we keep that discussion light and focus on the geometric tools that sit just below calculus.

In the 1800s, geometry gained an especially efficient algebraic interface: vector analysis, where the dot and cross products encode the two most basic geometric primitives--projection/angle and oriented area. The modern $\cdot$ and $\times$ notation was introduced by Gibbs (and independently Heaviside) around 1881 and then standardized in physics usage. Later, divergence and curl became standard because they compactly describe fields and their integral laws--but that differential layer is not our focus here.

That same "area backbone" explains the next conceptual step. Grassmann (1844) introduced the wedge product and exterior algebra, where the fundamental geometric object is no longer just a vector but also a bivector (directed area), trivector (directed volume), and so on. Clifford (1878) then built geometric algebra by unifying these objects into one coherent multiplication system. In our 2D Euclidean setting, $$|\mathbf{u}\times\mathbf{t}|,\qquad \|\mathbf{u}\wedge\mathbf{t}\|,\qquad |\omega(\mathbf{u},\mathbf{t})|$$ are simply three notations for the same invariant: the area spanned by $\mathbf{u}$ and the tangent direction $\mathbf{t}$. So the proofs are interchangeable across these languages with no change in geometric content.

Finally, modern physics is "geometric" in a deeper sense: it naturally lives on spaces that are not just 2D pictures--3D space, 4D spacetime, and even higher-dimensional configuration/phase spaces. We don't always call those "geometry" in everyday language because they're hard to visualize, but mathematically they are geometry: they carry intrinsic notions of angle, area/flux, invariance, and coordinate-free meaning. In our ellipse problem, we see a toy version of the same idea: at each point $E$ on the ellipse, there is a tangent line $DE$. Thinking of the tangent line as the attached object that moves with $E$ is the simplest concrete example of the bundle/fiber mindset.

A brief timeline of tooling relevant to this note:

  • ~300 BCE: Euclid -- synthetic geometry
  • 1637: Descartes -- analytic geometry
  • 1687: Newton -- Euclidean geometry + limiting arguments
  • 1827: Gauss -- intrinsic geometric invariants (surfaces)
  • 1844: Grassmann -- wedge / bivectors (area as an object)
  • 1878: Clifford -- geometric algebra (bivectors/trivectors as first-class)
  • 1881: Gibbs/Heaviside -- standard dot/cross notation
  • 1899: Cartan -- exterior derivative/forms (beyond our needs here)

For readers familiar with Maxwell's equations, the table below shows how mathematical language for geometry evolved over time: from coordinate-expanded component equations, to vector analysis, to geometric algebra, and then to differential forms. The tools change, but the physics does not, even as the expressions become more compact.

Formulation Coordinate status (what's really true) Maxwell equations written in that language What you gain / lose Good reference
Maxwell (1865, component-heavy) Coordinate-invariant laws, but written in components (explicit $x,y,z$ bookkeeping) and bundled with extra relations (constitutive laws, Ohm, forces, potentials). Not packaged as the modern four; a larger coupled component system (often summarized as ~20 equations/unknowns, plus auxiliaries). Gain: historically close to derivations. Lose: compactness; structure is obscured by bookkeeping. Maxwell's 1865 paper [MAXWELL-1865]; Deschamps [FORMS-EM].
Vector calculus (Heaviside/Gibbs) Coordinate-free in content; coordinates enter only when writing component formulas. (SI, vacuum) $\nabla\cdot\mathbf{E}=\rho/\varepsilon_0$
$\nabla\cdot\mathbf{B}=0$
$\nabla\times\mathbf{E}=-\partial_t\mathbf{B}$
$\nabla\times\mathbf{B}=\mu_0\mathbf{J}+\mu_0\varepsilon_0\,\partial_t\mathbf{E}$.		 Gain: four clean equations; practical in 3D Euclidean space. Caveat: $\nabla\cdot,\nabla\times$ depend on the chosen metric + orientation, though invariantly defined once fixed. Wilson [VECTOR-ANALYSIS]; cross product [CROSS-PRODUCT]; vector analysis [VECTOR-ANALYSIS-WIKI].
Geometric algebra (Clifford; dot + wedge unified) Coordinate-free once metric + orientation are fixed. Vectors are bold (e.g., $\mathbf{a},\mathbf{b}$); higher-grade objects (bivectors, trivectors) are not vectors in disguise. Core algebra: $\mathbf{a}\mathbf{b}=\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\wedge\mathbf{b}$. Maxwell can be written compactly (common spacetime GA convention): $\boxed{\nabla F=\mu_0 J}$, with the grade split $\nabla F=\nabla\cdot F+\nabla\wedge F$ giving $\nabla\wedge F=0$ (homogeneous) and $\nabla\cdot F=\mu_0 J$ (sources). Gain: cross product becomes derived (dual of $\wedge$); div/curl unify as one operator split by grade; rotations often cleaner. Lose: conventions vary (3D GA vs spacetime GA; units/signatures); less standard in mainstream EM courses. Peeter Joot [GA-EM].
Differential forms (exterior calculus on spacetime) Manifestly coordinate-invariant; works naturally on curved manifolds (GR). Metric enters via the Hodge star. $\boxed{dF=0}$, $\boxed{d{*}F=J}$. Here $*$ maps $k$-forms to $(n-k)$-forms using the metric + orientation (in 4D, $2\leftrightarrow2$). Gain: topology/relativity are transparent; $d^2=0$ exposes structure. Lose: abstraction up front; computations often unpack to components. Deschamps [FORMS-EM]; Lindell [LINDELL-DF]; Epstein [EPSTEIN-DG].

Table 1 summarizes the evolution of Maxwell's equations across these languages.

Notes:

  • Geometric algebra (Clifford; dot + wedge unified): metric + wedge fused immediately via $\mathbf{a}\mathbf{b}=\mathbf{a}\cdot\mathbf{b}+\mathbf{a}\wedge\mathbf{b}$.
  • Differential forms (exterior calculus on spacetime): forms are wedge-first, metric-later via $*$.

So the concept of vector analysis has evolved significantly since Newton's time. Maxwell's equations were first written in a coordinate-expanded form, which has the flavor of our solution in Part 2, along with differential operators that were by then well understood. Only later did divergence and curl become standard, giving a more compact, coordinate-free expression using $\nabla\cdot$ and $\nabla\times$. Those operators are coordinate-free in meaning, which parallels the shift from the coordinate solution in Part 2 back to the coordinate-free constructions in Part 1. Vector analysis keeps the geometric insight while still enabling the algebraic computations of the coordinate approach. To explain the power of this toolbox, I include a couple of short proofs below.

Since our problem is 2D, both dot and cross products can be used to solve it naturally. We emphasize how these tools enable coordinate-free proofs while still allowing analytic methods beyond the purely Euclidean geometry used in Part 1, yet more direct than the coordinate approach in Part 2.

Proof 4.1 (reflection on $DE$)¶

Use the reflection property: a ray from $B$ to $E$ reflects across $DE$ and passes through $H$. Let $p=|\overrightarrow{EB}|$, $q=|\overrightarrow{EH}|$, and define unit vectors $\hat u=\overrightarrow{EB}/p$, $\hat v=\overrightarrow{EH}/q$. Then $\overrightarrow{EB}=p\hat u$ and $\overrightarrow{EH}=q\hat v$, with $|\hat u|=|\hat v|=1$. The tangent at $E$ is perpendicular to $\hat u+\hat v$, so its direction is parallel to $\hat u-\hat v$. Hence $$\overrightarrow{ED}=\tau(\hat u-\hat v)$$ for some scalar $\tau$. Because $D\in BH$, $$\overrightarrow{BH}=\overrightarrow{EH}-\overrightarrow{EB}=q\hat v-p\hat u,\quad \overrightarrow{BD}=\mu\overrightarrow{BH}.$$ Using $\overrightarrow{ED}=\overrightarrow{EB}+\overrightarrow{BD}$ gives $$\tau(\hat u-\hat v)=p(1-\mu)\hat u+\mu q\hat v.$$ Matching coefficients yields $\tau=p(1-\mu)$ and $-\tau=\mu q$, so $\mu=\frac{p}{p-q}$. Finally, $\overrightarrow{HD}=\overrightarrow{BD}-\overrightarrow{BH}=(\mu-1)\overrightarrow{BH}$, so $$\frac{BD}{HD}=\frac{\mu}{\mu-1}=\frac{p}{q}=\frac{BE}{EH},$$ which is equivalent to $\frac{BE}{BD}=\frac{EH}{HD}$.

Proof 4.2 (using the cross product)¶

Let the tangent at $E$ meet $BH$ at $D$, and take any nonzero tangent vector $\overrightarrow{ED}$. The reflection law says the tangent bisects the angle between $\overrightarrow{EB}$ and $\overrightarrow{EH}$, so the perpendicular components match: $$\frac{|\overrightarrow{EH}\times\overrightarrow{ED}|}{EH}=\frac{|\overrightarrow{EB}\times\overrightarrow{ED}|}{EB}.$$ Now $\overrightarrow{EH}=\overrightarrow{ED}+\overrightarrow{DH}$ and $\overrightarrow{EB}=\overrightarrow{ED}+\overrightarrow{DB}$. Cross with $\overrightarrow{ED}$ and drop the $\overrightarrow{ED}\times\overrightarrow{ED}$ term: $$\overrightarrow{EH}\times\overrightarrow{ED}=\overrightarrow{DH}\times\overrightarrow{ED},\qquad \overrightarrow{EB}\times\overrightarrow{ED}=\overrightarrow{DB}\times\overrightarrow{ED}.$$ So $$\frac{|\overrightarrow{DH}\times\overrightarrow{ED}|}{EH}=\frac{|\overrightarrow{DB}\times\overrightarrow{ED}|}{BE}.$$ Because $B,D,H$ are collinear, $\overrightarrow{DB}\parallel\overrightarrow{DH}$, hence $$|\overrightarrow{DB}\times\overrightarrow{ED}|:|\overrightarrow{DH}\times\overrightarrow{ED}|=BD:HD.$$ Substitute to get $$\boxed{\frac{BD}{HD}=\frac{BE}{EH}},$$ which is equivalent to $\frac{BE}{BD}=\frac{EH}{HD}$.

I explored several alternative constructions which give various other proofs that I won't detail here since these two are the simplest and representative. For instance, in the original configuration, the condition $EH\perp AB$ immediately yields the dot-product constraint $\overrightarrow{EH}\cdot \overrightarrow{AB}=0$. This orthogonality can be used to reduce the degrees of freedom without appealing to any reflection argument. More generally, one can start from essentially any point in the diagram, parameterize the remaining configuration by a small set of length (or scaling) parameters, and then drive the argument to the target length-ratio identity. There are many such routes, and they do not require introducing any auxiliary lines.

In this 2D Euclidean setting, the geometric-algebra approach does not introduce additional structure compared with the cross product. Both encode the oriented (signed) area of the parallelogram spanned by two vectors. The $\times$ operator only makes traditional sense in 3D space (it produces a vector perpendicular to the plane), whereas the $\wedge$ operator gives a bivector that is defined in any dimension. Likewise, the differential-forms formulation is not fundamentally different here: a 2-form takes two vectors as input and returns a scalar, and a 1-form is essentially a linear functional (often represented by a dot product with a fixed vector). The identities therefore reduce to the same area or determinant computations already used in the cross-product argument. For completeness, Proof 4.3 shows how they are unified in one proof. In modern math/physics, where the dimensionality is often beyond 3, the dot/cross picture becomes awkward, which is why spacetime formulations separate time from space and why geometric algebra and differential forms are preferred. That is also why the unified Maxwell equations are usually written without an explicit time coordinate.

Proof 4.3: Unified simple solution (area language: cross = wedge = 2-form = 1-form)¶

Let $DE$ be the tangent at $E$ and let $\mathbf{t}=\overrightarrow{ED}\ne 0$. For any vector $\mathbf{u}$ based at $E$ (here $\overrightarrow{EB},\overrightarrow{EH},\overrightarrow{DB},\overrightarrow{DH}$), the area with the tangent is the same in all languages: $$A(\mathbf{u})=|\mathbf{u}\times\mathbf{t}|=\|\mathbf{u}\wedge\mathbf{t}\|=|\omega(\mathbf{u},\mathbf{t})|,$$ where $\omega$ is an oriented area 2-form.

Fix $\omega$ and define the 1-form $\alpha=\iota_{\mathbf{t}}\omega$, so $\alpha(\mathbf{u})=\omega(\mathbf{u},\mathbf{t})$ and $|\alpha(\mathbf{u})|=A(\mathbf{u})$. This is the perpendicular component of $\mathbf{u}$ to the tangent.

Reflection at $E$ gives equality of perpendicular components: $$\frac{A(\overrightarrow{EH})}{EH}=\frac{A(\overrightarrow{EB})}{EB}.$$

Since $D\in BH$, $$\overrightarrow{EH}=\overrightarrow{ED}+\overrightarrow{DH},\qquad \overrightarrow{EB}=\overrightarrow{ED}+\overrightarrow{DB}.$$ Using bilinearity and $\omega(\mathbf{t},\mathbf{t})=0$ gives $$A(\overrightarrow{EH})=A(\overrightarrow{DH}),\qquad A(\overrightarrow{EB})=A(\overrightarrow{DB}).$$ Hence $$\frac{A(\overrightarrow{DH})}{EH}=\frac{A(\overrightarrow{DB})}{EB}.$$

Because $B,D,H$ are collinear, $A(\overrightarrow{DB})/A(\overrightarrow{DH})=BD/HD$, so $$\frac{BD}{HD}=\frac{BE}{EH},\quad\text{i.e.}\quad \frac{BE}{BD}=\frac{EH}{HD}.$$

By now I hope this part of the note has conveyed an appreciation for coordinate-free methods in our mathematical toolbox. I also hope it gives a preview of what modern physics use to solve geometry problems that are more than 3-dimensional, admittedly in a weak sense.

So far in this part we have discussed the simple case of how the reflection property implies the ratio equality. Below we show why the original problem in Part 1 gives the reflection property of $EF$ and $BE$, using a vector argument.

Proof 4.4 Vector analysis: $EF$ is the reflection of $BE$ across $CD$ in Figure 3¶

Let $\mathbf{u}=\overrightarrow{CB}$ and $\mathbf{d}=\overrightarrow{CD}$. Then $\mathbf{u}\perp\mathbf{d}$ and $\lVert\mathbf{u}\rVert=\lVert\mathbf{d}\rVert$. Fix the unit normal $\mathbf{w}=\dfrac{\mathbf{u}\times\mathbf{d}}{\lVert\mathbf{u}\times\mathbf{d}\rVert}$. Since $AD\parallel CB$, $E\in CD$, and $DE=AD$, there exists a scalar $k$ such that $$\overrightarrow{AB}=-\mathbf{d}+(1-k)\mathbf{u},\qquad \overrightarrow{BE}=-(k-1)\mathbf{d}-\mathbf{u}. $$ Reflection across $CD$ is equivalent to $$\frac{\lVert\overrightarrow{BE}\times\mathbf{d}\rVert}{\lVert\overrightarrow{BE}\rVert}=\frac{\lVert\overrightarrow{EF}\times\mathbf{d}\rVert}{\lVert\overrightarrow{EF}\rVert}. $$ Because $EF\perp AB$, the direction of $EF$ is parallel to $\overrightarrow{AB}\times\mathbf{w}$, so the right-hand side equals $$\frac{\lVert(\overrightarrow{AB}\times\mathbf{w})\times\mathbf{d}\rVert}{\lVert\overrightarrow{AB}\times\mathbf{w}\rVert}. $$ Now compute the two normalized quantities. First, $$\overrightarrow{BE}\times\mathbf{d}=(-(k-1)\mathbf{d}-\mathbf{u})\times\mathbf{d}=-\mathbf{u}\times\mathbf{d}, $$ so $\lVert\overrightarrow{BE}\times\mathbf{d}\rVert=\lVert\mathbf{u}\times\mathbf{d}\rVert=\lVert\mathbf{u}\rVert\,\lVert\mathbf{d}\rVert=\lVert\mathbf{d}\rVert^2$. Next, by Lagrange's formula $(\mathbf{a}\times\mathbf{b})\times\mathbf{c}=\mathbf{b}(\mathbf{a}\cdot\mathbf{c})-\mathbf{a}(\mathbf{b}\cdot\mathbf{c})$ and $\mathbf{w}\cdot\mathbf{d}=0$, $$ (\overrightarrow{AB}\times\mathbf{w})\times\mathbf{d}=\mathbf{w}(\overrightarrow{AB}\cdot\mathbf{d}). $$ But $$\overrightarrow{AB}\cdot\mathbf{d}=(-\mathbf{d}+(1-k)\mathbf{u})\cdot\mathbf{d}=-\lVert\mathbf{d}\rVert^2, $$ hence $\lVert(\overrightarrow{AB}\times\mathbf{w})\times\mathbf{d}\rVert=\lVert\mathbf{d}\rVert^2$. Finally, $\lVert\overrightarrow{AB}\times\mathbf{w}\rVert=\lVert\overrightarrow{AB}\rVert$, and $$\lVert\overrightarrow{AB}\rVert^2=\lVert\mathbf{d}\rVert^2+(k-1)^2\lVert\mathbf{u}\rVert^2=(k-1)^2\lVert\mathbf{d}\rVert^2+\lVert\mathbf{u}\rVert^2=\lVert\overrightarrow{BE}\rVert^2, $$ so $\lVert\overrightarrow{AB}\rVert=\lVert\overrightarrow{BE}\rVert$. Therefore the normalized cross magnitudes match, and the reflection condition holds: $$\frac{\lVert\overrightarrow{BE}\times\overrightarrow{CD}\rVert}{\lVert\overrightarrow{BE}\rVert}=\frac{\lVert\overrightarrow{EF}\times\overrightarrow{CD}\rVert}{\lVert\overrightarrow{EF}\rVert}. $$ So $BE$ reflects across $CD$ into $EF$.

We used Lagrange’s formula (the vector triple product identity) in the proof above. There are many known proofs, some classic and purely algebraic, but this one that I came up with is purely geometric and makes the identity feel almost unavoidable. I haven’t seen this particular “Thales-circle + chord” intuition presented in such a compact form elsewhere, so it seems worth sharing though it can hardly be original, given that this is elementary. In a sense, it brings the argument back to a Euclid-style picture: analysis and geometry are not rivals, but two languages for the same structure, constantly translating into each other.

Proof of Lagrange's formula based on only geometric properties¶

Let $O$ be the origin and $\overrightarrow{OU}=\mathbf{u}$, $\overrightarrow{OV}=\mathbf{v}$, $\overrightarrow{OW}=\mathbf{w}$. Write $\mathbf{w}=\mathbf{w}_{\parallel}+\mathbf{w}_{\perp}$ relative to the plane $(OU,OV)$. Then $(\mathbf{u}\times\mathbf{v})\times\mathbf{w}_{\perp}=0$ (since $\mathbf{u}\times\mathbf{v}\parallel\mathbf{w}_{\perp}$) and $\mathbf{u}\cdot\mathbf{w}_{\perp}=\mathbf{v}\cdot\mathbf{w}_{\perp}=0$, so it suffices to assume $U,V,W$ are coplanar. By linearity of both $\times$ and $\cdot$ in each argument, scaling $\mathbf{u},\mathbf{v},\mathbf{w}$ by $a,b,c$ multiplies both sides by $abc$; hence it suffices to prove the unit case $\lvert\mathbf{u}\rvert=\lvert\mathbf{v}\rvert=\lvert\mathbf{w}\rvert=1$. Let $\alpha=\angle UOV$.

Image

Figure 11: Geometric property of the triple vector formula.

For the plane geometry, refer to Figure 11. Draw the circle with diameter $OW$. Let it meet rays $OU,OV$ again at $A,B$. By Thales, $\angle OAW=\angle OBW=90^\circ$, hence $OA=\mathbf{u}\cdot\mathbf{w}$ and $OB=\mathbf{v}\cdot\mathbf{w}$. Define $A'\in OV$ and $B'\in OU$ by $OA'=OA$ and $OB'=OB$. Then $$\overrightarrow{A'B'}=(\mathbf{v}\cdot\mathbf{w})\mathbf{u}-(\mathbf{u}\cdot\mathbf{w})\mathbf{v}= -\big((\mathbf{u}\cdot\mathbf{w})\mathbf{v}-(\mathbf{v}\cdot\mathbf{w})\mathbf{u}\big),$$ so $\lvert A'B'\rvert$ is the magnitude of the right-hand side. It is easy to show $A'B'\perp OW$ by checking $\overrightarrow{A'B'}\cdot\mathbf{w}=0$, but here we want a pure geometric proof. Since $\angle WOA + \angle A'B'O=\angle WBA + \angle ABO=90^\circ$, we get $A'B'\perp OW$, matching the left-hand side direction. Since $\triangle OAB\cong\triangle OA'B'$, we have $AB=A'B'$. In the diameter-1 circle, $AB=\sin\alpha$. But $$\lvert(\mathbf{u}\times\mathbf{v})\times\mathbf{w}\rvert=\lvert\mathbf{u}\times\mathbf{v}\rvert\,\lvert\mathbf{w}\rvert=\sin\alpha,$$ so both sides are the same vector: $$ (\mathbf{u}\times\mathbf{v})\times\mathbf{w}=(\mathbf{u}\cdot\mathbf{w})\mathbf{v}-(\mathbf{v}\cdot\mathbf{w})\mathbf{u}. $$ In short, we used the multilinearity of the dot and cross products (it is enough to prove the identity for coplanar unit vectors). We also used that dot and cross magnitudes correspond to cos and sin of the angle. The rest relies on a circle fact: two chords have the same length if and only if their inscribed angles are equal.

Postscript¶

It can be interesting to see how the following software would approach the problem:

  1. Medal-level AI geometry solvers (closest to "gold/silver" performance in the International Math Olympiad)
  • AlphaGeometry / AlphaGeometry 2 (research systems that can solve many Olympiad problems)
  • Open-source software such as Newclid/Yuclid by the startup company Harmonic
  1. Geometry theorem provers (strong automation, not always Olympiad-style proofs)
  • GCLC / WinGCLC (area method, Wu's method, Gr?bner bases)
  • JGEX and SymPy (Wu/Grobner/full-angle methods with visualization)
  • OpenGeoProver (Wu/Gr?bner style)
  • GeoLogic (interactive Euclidean prover)
  1. Proof assistants (formal, but manual)
  • Lean 4 + mathlib
  • LeanGeo

References¶

  • [CONICS-WIKIPEDIA] "Conic section." Wikipedia. Summary: Overview of conic sections and their standard definitions. URL: https://en.wikipedia.org/wiki/Conic_section
  • [CROSS-PRODUCT] Gibbs/Heaviside (1881). "Cross product." Summary: Notation for dot/cross products introduced in 1881. URL: https://en.wikipedia.org/wiki/Cross_product
  • [EPSTEIN-DG] Epstein (2013~2014). "Differential Geometry Minicourse." Summary: Minicourse notes associated with a 2013 workshop; book version appeared in 2014. URL: https://marceloepstein.weebly.com/uploads/1/2/8/4/12849112/dg_minicourse.pdf (context: https://www.isbn.de/buch/9783319357140/differential-geometry)
  • [EUCLID-VIDEO] Syversen (2025-09-18). "Euclid's ruler-and-compass constructions." YouTube. Summary: Video on the role of straightedge-and-compass constructions in Euclid. URL: https://youtu.be/M-MgQC6z3VU
  • [FORMS-EM] Deschamps (1981). "Electromagnetics and Differential Forms." Summary: Proc. IEEE article that sparked broader EM use of differential forms. URL: https://faculty.washington.edu/seattle/physics544/2011-lectures/deschamps.pdf
  • [GA-EM] Joot (2018). "Geometric Algebra for Electrical Engineers: Multivector Electromagnetism." Summary: Notes introducing geometric algebra for electromagnetism (version dated April 2018). URL: https://peeterjoot.com/archives/math2018/GAelectrodynamics.V0.1.6.pdf
  • [KEPLER-TEACH] van Haandel & Heckman (2007). "Teaching the Kepler laws for freshmen." arXiv:0707.4605. Summary: Pedagogical treatment of Kepler's laws with geometric insights. URL: https://arxiv.org/abs/0707.4605
  • [LINDELL-DF] Lindell (2004). "Differential Forms in Electromagnetics." Summary: Lecture notes/book with a 2004 copyright. URL: https://faculty.washington.edu/seattle/physics544/2011-lectures/lindell.pdf
  • [MAXWELL-1865] Maxwell (1865). "A Dynamical Theory of the Electromagnetic Field." Summary: Original formulation of Maxwell's equations in component-heavy form. URL: https://archive.org/details/electromagnetic00maxwgoog
  • [NEWTON-PRINCIPIA] Newton (1687). Philosophiae Naturalis Principia Mathematica. Project Gutenberg. Summary: Foundational work in classical mechanics with Euclidean-style geometric reasoning. URL: https://www.gutenberg.org/ebooks/28233
  • [QUATERNION] Hamilton (1843). "Quaternion." Summary: Quaternions were introduced by Hamilton in 1843. URL: https://en.wikipedia.org/wiki/Quaternion
  • [VECTOR-ANALYSIS] Wilson (1901). Vector Analysis (from Gibbs' lectures). Summary: Classic text that standardizes dot/cross notation and vector calculus. URL: https://archive.org/details/vectoranalysis00wilsrich
  • [VECTOR-ANALYSIS-WIKI] Wilson/Gibbs (1901). "Vector Analysis." Summary: Wikipedia entry on the 1901 textbook and its role in standardizing notation. URL: https://en.wikipedia.org/wiki/Vector_Analysis