JBMO Shortlist

Geometry Difficulty 4.3 AIME Prove it JBMO

Problem:

Let $ABCD$ be a trapezoid with $AB \parallel CD$ and $AB > CD$ . It is given that the sum of the angles at the base $AB$ is equal to $90^{\circ}$ . Prove that the distance between the midpoints of the parallel sides is equal to $\frac{1}{2}(AB - CD)$ .

Solution

Solution:

Let $E$ be the intersection point of $AD$ and $BC$ . The point $E$ is on the line through the midpoints $K$ and $F$ of the parallel sides. Let $G$ and $H$ be the points on $AB$ , such that $EG$ and $CH$ are parallel to $AE$ . Then the triangle $CHG$ is right-angled with the right angle at $C$ and the median $CO$ , $SO$ .

$KF = CG = \frac{1}{2}HB = \frac{1}{2}(AB - AH) = \frac{1}{2}(AB - CD)$

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