Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.9 AIME Prove it Austria

Let $ABCD$ be a trapezoid with parallel sides $AB$ and $CD$ , with $\angle BAD = 90^\circ$ and with $AB + CD = BC$ . Furthermore, let $M$ be the mid-point of $AD$ .
Prove that $\angle CMB = 90^\circ$ .

Solution

We reflect the points $B$ and $C$ in $M$ and obtain the points $E$ and $F$ , respectively. We clearly have $EC = BF = AB + AF = AB + CD = BC = EF$ , therefore, the quadrilateral $BCEF$ is a rhombus. Since the diagonals in a rhombus are orthogonal, we get $BE \perp CF$ and we obtain $\angle BMC = 90^\circ$ as desired.

Figure 1: Problem 2

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