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Berkeley Math Circle: Monthly Contest 2

Geometry Difficulty 4.4 AIME Prove it United States

Problem:

Let $ABCD$ be an isosceles trapezoid, and let $E$ be the foot of the altitude from $A$ to line $BC$ . Prove that line $DE$ passes through the centroid of $\triangle ABC$ .

Solution

Solution:

Let $M$ be the midpoint of $BC$ . Also, let $F$ be the foot from $D$ to $BC$ . Then $AEFD$ is a rectangle. Define $G$ as the intersection of $AM$ and $DE$ . Then $\triangle AGD \sim \triangle EMG$ , and we get

$\frac{GM}{GA} = \frac{ME}{DA} = \frac{ME}{FE} = \frac{1}{2}$

which implies $G$ is the centroid of $\triangle ABC$ , since the centroid divides the $A$ -median in a $2:1$ ratio. Thus $G$ lies on $DE$ .

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