bad solution but I coord bashed this: A(0,0) B(0,y) C(x,y) D(x,0) $P=(\frac{x^3}{x^2 + y^2},\frac{x^2 y}{x^2 +y^2})$. Now we just have to confirm that it is the midpoint of AP satisfying this M(x/2, y) Let N be the midpoint of AP $N=(\frac{x^3}{2(x^2 + y^2)},\frac{x^2 y}{2(x^2 +y^2)})$ Now clearly there is only one point on the line segment AP such that <DNM=90 degrees The slope of MN is $\frac{x^2 +2y^2}{xy}$ and the slope of ND is $\frac{-xy}{x^2 + 2y^2}$ so the product of the slopes is -1(perpendicular/angle is 90 degrees). Thus N=Q.

A more conventional solution: Define instead $Q$ to be the midpoint, and we show $\angle DQM$ is right. It suffices to show $DCMQ$ cyclic. Since $\triangle DAP \sim \triangle DBC$ we have by the midpoints that $\triangle DQP \sim \triangle DMC$, so $\angle DQC = \angle DMC$, QED.

Direct solution: 1) $m(\widehat{CDM})=m(\widehat{BAM})$ by symmetry. 2) $m(\widehat{CDM})=m(\widehat{QDP})$ as $DP, DM$ being isogonal lines inside $\angle CDQ$ (altitude and diameter of the circumcircle) 3) Consequently $m(\widehat{QDP})=m(\widehat{BAM})$ and, since triangles $ADP$ and $CAB$ are similar, we infer the points $M, Q$ being corresponding (dividing segments BC, AP respectively at the same ratio), i.e., if $M$ is midpoint of BC, Q must be midpoint of $AP$. Best regards, sunken rock

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