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Geometry Difficulty 4.7 AIME Prove it JBMO

Problem:

In a right trapezoid $ABCD$ ( $AB \parallel CD$ ) the angle at vertex $B$ measures $75^{\circ}$ . Point $H$ is the foot of the perpendicular from point $A$ to the line $BC$ . If $BH = DC$ and $AD + AH = 8$ , find the area of $ABCD$ .

Solution

Solution:

Produce the legs of the trapezoid until they intersect at point $E$ . The triangles $ABH$ and $ECD$ are congruent (ASA). The area of $ABCD$ is equal to area of triangle $EAH$ of hypotenuse
$AE = AD + DE = AD + AH = 8$
Let $M$ be the midpoint of $AE$ . Then
$ME = MA = MH = 4$
and $\angle AMH = 30^{\circ}$ . Now, the altitude from $H$ to $AM$ equals one half of $MH$ , namely $2$ . Finally, the area is $8$ .

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