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Olimpiada Matemática del Cono Sur

Geometry Difficulty 5.1 AIME, harder Prove it Argentina

In a convex quadrilateral $ABCD$ we have that:
* $R$ and $S$ are points in the interior of the segments $CD$ and $AB$ respectively, with $AD = CR$ and $BC = AS$ .
* $P$ and $Q$ are the midpoints of $DR$ and $SB$ respectively.
* $M$ is the midpoint of $AC$ .
If it is known that $MPC + MQA = 90°$ , prove that $ABCD$ is a cyclic quadrilateral.

Solution

Let $E$ and $F$ be points in the prolongations of $AB$ and $CD$ such that $DF = DA = CR$ and $BE = BC = AS$ , as in the picture.

Note that $P$ is the midpoint of $FC$ . Then, $MP$ is a midsegment of the triangle $AFC$ , which implies that $A\hat{F}C = M\hat{P}C$ . Since the triangle $ADF$ is isosceles, we have that $D\hat{A}F = D\hat{F}A = M\hat{P}C$ , and then, $A\hat{D}C = 2 \cdot M\hat{P}C$ (external angle).
Similarly, $A\hat{B}C = 2 \cdot M\hat{Q}A$ .
Thus,
$A\hat{D}C + A\hat{B}C = 2 \cdot (M\hat{P}C + M\hat{Q}A) = 2 \cdot 90^\circ = 180^\circ,$
and, therefore, the quadrilateral $ABCD$ is cyclic.

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