perpendicularity in an inscribed ABCD, BC=DC, AB =AC

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perpendicularity in an inscribed ABCD, BC=DC, AB =AC

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Source: All-Siberian Open School Olympiad 2018-19 10.5

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#1 h Jul 6, 2020, 6:57 AM

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The quadrilateral $ABCD$ is inscribed in a circle with $BC=DC$ and $AB =AC$ . Let the point $P$ be the midpoint of the arc $CD$ not containing point $A$ , and $Q$ is the intersection point of the diagonals $AC$ and $BD$ . Prove that lines $PQ$ and $AB$ are perpendicular.

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#2 h Jul 6, 2020, 3:18 PM

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My solution.

Let $\omega$ is the circle through $A$ , $B$ , $C$ and $D$ .
Let $O$ be centre of the circle $\omega$ . Let line $AO$ and circle $\omega$ intersect at $F$ . Thus, $AF$ is a diameter of the circle $\omega$ .
Let $G$ be intersection point of $PB$ and $DF$ .
Now we only need to show that $PQ \parallel BF$ .
We see that $F$ is defined similarly as $P$ , so $DP=PC=CF=BF$ . Furthermore, we get that $PGFC$ is a rhombus and thus $PB\perp AC$ and thus $QB=BC$ and thus $PQ=PC=FB$ and thus $PQBF$ is a parallelogram and hence, we have shown that $PQ \parallel BF$ .

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