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Saudi Arabian Mathematical Competitions

Geometry Difficulty 4.8 AIME Prove it Saudi Arabia

In square $ABCD$ with side $1$ , point $E$ lies on $BC$ and $F$ lies on $CD$ such that $\angle EAB = 20^{\circ}$ , $\angle EAF = 45^{\circ}$ . Find the length of altitude $AH$ of $\triangle AEF$ .

Solution

Take point $G$ on the opposite ray of ray $DC$ such that $GD = BE$ . Then two triangles $ABE$ , $ADG$ are congruent, implies that $AE = AG$ and $\angle EAB = \angle DAG = 20^{\circ}$ . Thus
$\angle FAG = \angle FAD + \angle DAG = \angle EAB + \angle DAG = 45^{\circ}.$
Thus two triangles $AEF$ , $AGF$ have common side $AF$ and $AE = AG$ and $\angle FAG = \angle FAE = 45^{\circ}$ , thus $\triangle AEF \cong \triangle AGF$ , then $AH = AD = 1$ , where $AH$ is the altitude with respect to vertex $A$ in triangle $AEF$ .

$\square$

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