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Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.0 AIME Prove it United States

Problem:

$ABCD$ is a convex quadrilateral such that $AB < AD$ . The diagonal $\overline{AC}$ bisects $\angle BAD$ , and $m \angle ABD = 130^{\circ}$ . Let $E$ be a point on the interior of $\overline{AD}$ , and $m \angle BAD = 40^{\circ}$ . Given that $BC = CD = DE$ , determine $m \angle ACE$ in degrees.

Solution

Solution:

First, we check that $ABCD$ is cyclic. Reflect $B$ over $\overline{AC}$ to $B'$ on $\overline{AD}$ , and note that $B'C = CD$ . Therefore, $m \angle ADC = m \angle B'DC = m \angle CB'D = 180^{\circ} - m \angle AB'C = 180^{\circ} - m \angle CBA$ .

Now $m \angle CBD = m \angle CAD = 20^{\circ}$ and $m \angle ADC = 180^{\circ} - m \angle CBA = 30^{\circ}$ . Triangle $CDE$ is isosceles, so $m \angle CED = 75^{\circ}$ and $m \angle AEC = 105^{\circ}$ . It follows that $m \angle ECA = 180^{\circ} - m \angle AEC - m \angle CAE = 55^{\circ}$ .

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