NZMO Round One

Geometry Difficulty 5.1 AIME, harder Prove it New Zealand

Problem:
Let $ABCD$ be a trapezium such that $AB \parallel CD$ . Let $E$ be the intersection of diagonals $AC$ and $BD$ . Suppose that $AB = BE$ and $AC = DE$ . Prove that the internal angle bisector of $\angle BAC$ is perpendicular to $AD$ .

Solution

Solution:
First note that triangle $ABE$ is isosceles because $AB = BE$ .

Let $x = \angle DEC$ . Angle chasing gives:
$x = \angle DEC = \angle BEA = \angle EAB = \angle ACD.$
Therefore triangle $CDE$ is isosceles. Hence $DE = DC$ . Since we are also given $AC = DE$ this implies $AC = DC$ . Therefore $\triangle ACD$ is isosceles. Since $x = \angle ACD$ this gives us
$\angle CDA = \angle DAC = 90^\circ - \frac{x}{2}$
Now let $\lambda$ be the angle bisector of $\angle BAC$ . Since $x = \angle BAC$ we know that $\lambda$ makes an angle of $\frac{1}{2} \angle BAC = \frac{x}{2}$ with line $AC$ . Therefore the angle between $\lambda$ and $AD$ is
$\frac{x}{2} + \angle DAC = \frac{x}{2} + \left(90^\circ - \frac{x}{2}\right) = 90^\circ$
as required.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.