IRL ABooklet

Geometry Difficulty 6.5 National Olympiad Prove it Ireland

We are given a triangle ABC such that $\angle BAC < 90^\circ$ . The point $D$ is on the opposite side of the line $AB$ to $C$ such that $|AD| = |BD|$ and $\angle ADB = 90^\circ$ . Similarly, the point $E$ is on the opposite side of $AC$ to $B$ such that $|AE| = |CE|$ and $\angle AEC = 90^\circ$ . The point $X$ is such that $ADXE$ is a parallelogram. Prove that $|BX| = |CX|$ .

Solutions — 2

Solution 1

Since $ADXE$ is a parallelogram, we have $\angle ADX = \angle AEX$ . This implies that $\angle XDB = 90^\circ - \angle ADX = 90^\circ - \angle AEX = \angle CEX$ . Since triangle $ADB$ is isosceles and $ADXE$ is a parallelogram, we have $|DB| = |DA| = |XE|$ . Similarly, $|EC| = |EA| = |XD|$ . We conclude that triangles $XDB$ and $CEX$ are congruent by SAS and so $|BX| = |CX|$ .

Solution 2

First we note that $\angle DBA = \angle DAB = \angle ECA = \angle EAC = 45^\circ$ . Since $ADXE$ is a parallelogram, we have $\angle ADX + \angle DAE = 180^\circ$ . Thus
$\begin{align*} \angle BDX &= 90^\circ - \angle ADX = 90^\circ - (180^\circ - \angle DAE) = \angle DAE - 90^\circ \\ &= \angle DAE - (45^\circ + 45^\circ) = \angle DAE - \angle DAB - \angle EAC \\ &= \angle BAC. \end{align*}$
Next, note that $\frac{|AB|}{|BD|} = \sqrt{2} = \frac{|AC|}{|AE|} = \frac{|AC|}{|DX|}$ . Thus $\frac{|BD|}{|DX|} = \frac{|AB|}{|AC|}$ .
Combining $\angle BDX = \angle BAC$ and $\frac{|BD|}{|DX|} = \frac{|AB|}{|AC|}$ , we deduce that triangles $DBX$ and $ABC$ are similar. It follows that $\angle DBX = \angle ABC$ and so $\angle XBC = \angle DBA = 45^\circ$ . A similar argument shows that $\angle XCB = 45^\circ$ . Thus the triangle $XBC$ is isosceles, and $|BX| = |CX|$ .

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