Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it

-、(50 points) As shown in Figure 6, in the convex quadrilateral $ABCD$ , $M$ is the midpoint of side $AB$ , and $MC=MD$ . Perpendiculars are drawn from points $C$ and $D$ to sides $BC$ and $AD$ , respectively, and the intersection of these two perpendiculars is point $P$ . A perpendicular $PQ$ is drawn from point $P$ to $AB$ at $Q$ . Prove that $\angle PQC = \angle PQD$ .

Solution

As shown in Figure 11, connect $PA$ and $PB$ , and take the midpoints $E$ and $F$ of $PA$ and $PB$ respectively. Connect $EM$ , $ED$ , $FM$ , and $FC$ . Then quadrilateral $PEMF$ is a parallelogram. Thus, $\angle PEM = \angle PFM$ . Since $ME = \frac{1}{2} BP = CF$ , $MF = \frac{1}{2} AP = DE$ ,
$\begin{aligned} MD & = MC \\ & \Rightarrow \triangle DEM \cong \triangle MFC \\ & \Rightarrow \angle DEM = \angle MFC \\ \Rightarrow & \angle PED = \angle DEM - \angle PEM \\ & = \angle MFC - \angle PFM = \angle PFC \end{aligned}$
$\text{Also, } \angle PED = 2 \angle PAD, \angle PFC = 2 \angle PBC \text{, }$

we get $\angle PAD = \angle PBC$ .
Since $\angle PQA = \angle PDA = 90^{\circ}$ ,
$\angle PQB = \angle PCB = 90^{\circ} \text{, }$

then $P$ , $Q$ , $A$ , $D$ and $P$ , $Q$ , $B$ , $C$ are concyclic respectively. Therefore, $\angle PQD = \angle PAD$ , $\angle PQC = \angle PBC$ . Hence, $\angle PQC = \angle PQD$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.