Kvant Math Problem 198

The conditions place $H$ on the line $AB$ and $K$ on the line $BC$.

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Problem

Given a parallelogram $ABCD$. On the lines $AB$ and $BC$, points $H$ and $K$ are chosen, respectively, such that the triangles $KAB$ and $HCB$ are isosceles ($KA=AB$ and $HC=CB$). Prove that the triangle $KDH$ is also isosceles.

V. L. Gutenmacher

Exploration

The conditions place $H$ on the line $AB$ and $K$ on the line $BC$. Since $KA=AB$, triangle $KAB$ is isosceles with equal sides meeting at $A$. Since $HC=CB$, triangle $HCB$ is isosceles with equal sides meeting at $C$.

A coordinate model is tempting. Let

$$B=(0,0),\qquad A=\mathbf u,\qquad C=\mathbf v,\qquad D=\mathbf u+\mathbf v.$$

Because $H$ lies on the line $AB$, write $H=t\mathbf u$. The condition

$$HC=CB$$

gives

$$|t\mathbf u-\mathbf v|=|\mathbf v|.$$

After squaring,

$$t^2|\mathbf u|^2-2t(\mathbf u\cdot\mathbf v)=0,$$

hence

$$t=\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf u|^2}.$$

Similarly, since $K$ lies on the line $BC$, write $K=s\mathbf v$. The condition

$$KA=AB$$

gives

$$|s\mathbf v-\mathbf u|=|\mathbf u|,$$

so

$$s^2|\mathbf v|^2-2s(\mathbf u\cdot\mathbf v)=0,$$

hence

$$s=\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf v|^2}.$$

Now compute

$$D-H=\mathbf v+(1-t)\mathbf u,$$

and

$$D-K=\mathbf u+(1-s)\mathbf v.$$

The natural conjecture is that $DH=DK$. Expanding both squared lengths gives

$$DH^2=|\mathbf v|^2+(1-t)^2|\mathbf u|^2 +2(1-t)(\mathbf u\cdot\mathbf v),$$

$$DK^2=|\mathbf u|^2+(1-s)^2|\mathbf v|^2 +2(1-s)(\mathbf u\cdot\mathbf v).$$

Substituting the formulas for $s$ and $t$ causes all mixed terms to cancel, leaving

$$DH^2=|\mathbf u|^2+|\mathbf v|^2 -2(\mathbf u\cdot\mathbf v),$$

$$DK^2=|\mathbf u|^2+|\mathbf v|^2 -2(\mathbf u\cdot\mathbf v).$$

Thus $DH=DK$. The delicate point is the derivation of the parameters $s$ and $t$ from the isosceles conditions. Once those are obtained correctly, the final identity is straightforward.

Problem Understanding

We are given a parallelogram $ABCD$. A point $H$ is chosen on the line $AB$ so that $HC=CB$, and a point $K$ is chosen on the line $BC$ so that $KA=AB$. We must prove that triangle $KDH$ is isosceles.

This is a Type B problem.

The core difficulty is to translate the two isosceles conditions into information relating the positions of $H$ and $K$, and then show that two sides of triangle $KDH$ are equal.

Proof Architecture

Represent the parallelogram by vectors $A=\mathbf u$, $B=\mathbf0$, $C=\mathbf v$, $D=\mathbf u+\mathbf v$; this converts all geometric conditions into algebraic equalities of vectors.

Express $H$ as $t\mathbf u$ because $H$ lies on the line $AB$, and derive

$$t=\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf u|^2}$$

from the condition $HC=CB$.

Express $K$ as $s\mathbf v$ because $K$ lies on the line $BC$, and derive

$$s=\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf v|^2}$$

from the condition $KA=AB$.

Compute $DH^2$ and $DK^2$ in terms of $\mathbf u$, $\mathbf v$, $s$, and $t$.

Substitute the formulas for $s$ and $t$ and simplify to obtain $DH^2=DK^2$.

The most delicate lemma is the derivation of the parameters $s$ and $t$ from the equal side conditions.

Solution

Let

$$B=\mathbf0,\qquad A=\mathbf u,\qquad C=\mathbf v.$$

Since $ABCD$ is a parallelogram,

$$D=\mathbf u+\mathbf v.$$

Because $H$ lies on the line $AB$, there exists a real number $t$ such that

$$H=t\mathbf u.$$

The condition $HC=CB$ gives

$$|t\mathbf u-\mathbf v|=|\mathbf v|.$$

Squaring both sides,

$$t^2|\mathbf u|^2-2t(\mathbf u\cdot\mathbf v)+|\mathbf v|^2=|\mathbf v|^2.$$

Hence

$$t^2|\mathbf u|^2-2t(\mathbf u\cdot\mathbf v)=0.$$

The point $H=B$ corresponds to the trivial root $t=0$; the chosen point satisfying the isosceles condition is determined by

$$t=\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf u|^2}.$$

Because $K$ lies on the line $BC$, there exists a real number $s$ such that

$$K=s\mathbf v.$$

The condition $KA=AB$ gives

$$|s\mathbf v-\mathbf u|=|\mathbf u|.$$

Squaring,

$$s^2|\mathbf v|^2-2s(\mathbf u\cdot\mathbf v)+|\mathbf u|^2=|\mathbf u|^2,$$

hence

$$s^2|\mathbf v|^2-2s(\mathbf u\cdot\mathbf v)=0.$$

The nontrivial root is

$$s=\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf v|^2}.$$

Now

$$D-H=\mathbf v+(1-t)\mathbf u,$$

so

$$DH^2 =|\mathbf v|^2+(1-t)^2|\mathbf u|^2 +2(1-t)(\mathbf u\cdot\mathbf v).$$

Substituting

$$t=\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf u|^2},$$

we obtain

$$\begin{aligned} DH^2 &=|\mathbf v|^2 +\left(1-\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf u|^2}\right)^2|\mathbf u|^2 \ &\qquad +2\left(1-\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf u|^2}\right) (\mathbf u\cdot\mathbf v). \end{aligned}$$

After expansion,

$$DH^2 =|\mathbf u|^2+|\mathbf v|^2-2(\mathbf u\cdot\mathbf v).$$

Similarly,

$$D-K=\mathbf u+(1-s)\mathbf v,$$

hence

$$DK^2 =|\mathbf u|^2+(1-s)^2|\mathbf v|^2 +2(1-s)(\mathbf u\cdot\mathbf v).$$

Substituting

$$s=\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf v|^2},$$

gives

$$DK^2 =|\mathbf u|^2+|\mathbf v|^2-2(\mathbf u\cdot\mathbf v).$$

Therefore

$$DH^2=DK^2,$$

and consequently

$$DH=DK.$$

Thus triangle $KDH$ is isosceles. This completes the proof.

∎

Verification of Key Steps

The first delicate step is solving the condition $HC=CB$. Starting from

$$|t\mathbf u-\mathbf v|^2=|\mathbf v|^2,$$

one must expand both sides completely:

$$t^2|\mathbf u|^2-2t(\mathbf u\cdot\mathbf v)+|\mathbf v|^2=|\mathbf v|^2.$$

Cancelling $|\mathbf v|^2$ yields

$$t\bigl(t|\mathbf u|^2-2(\mathbf u\cdot\mathbf v)\bigr)=0.$$

Dropping the factor $t$ without noticing the root $t=0$ would be incorrect. The second root is precisely

$$t=\frac{2(\mathbf u\cdot\mathbf v)}{|\mathbf u|^2}.$$

The second delicate step is the simplification of $DH^2$. Writing $m=\mathbf u\cdot\mathbf v$,

$$DH^2 =|\mathbf v|^2+\left(1-\frac{2m}{|\mathbf u|^2}\right)^2|\mathbf u|^2 +2\left(1-\frac{2m}{|\mathbf u|^2}\right)m.$$

Expanding gives

$$|\mathbf v|^2+|\mathbf u|^2-4m+\frac{4m^2}{|\mathbf u|^2} +2m-\frac{4m^2}{|\mathbf u|^2},$$

and the quadratic terms cancel exactly, leaving

$$|\mathbf u|^2+|\mathbf v|^2-2m.$$

The third delicate step is recognizing that the same expression arises for $DK^2$. Any asymmetry in the calculation would signal an algebraic mistake, because the formulas for $s$ and $t$ are obtained from perfectly symmetric conditions.

Alternative Approaches

A synthetic proof can be obtained using rotations. Let $R_A$ be the rotation about $A$ sending $B$ to $K$, and let $R_C$ be the rotation about $C$ sending $B$ to $H$. Since $AB=AK$ and $CB=CH$, such rotations exist. Because opposite sides of a parallelogram are equal and parallel, the vectors involved in these two rotations are closely related. Composing suitable rotations shows that the images of $D$ under the two constructions lie at equal distances from $H$ and $K$, yielding $DH=DK$.

The vector method is preferable because it converts the two isosceles conditions into linear parameters on the sides of the parallelogram and leads directly to the identity $DH^2=DK^2$ with no auxiliary constructions.