A trapezoid with bases $AB$ and $CD$ is inscribed into a circle centered at $O$. Let $AP$ and $AQ$ be the tangents from $A$ to the circumcircle of triangle $CDO$. Prove that the circumcircle of triangle $APQ$ passes through the midpoint of $AB$.
Problem
Source: Sharygin 2019 finals Day 1 Grade 8 P1
Tags: geometry, Sharygin Geometry Olympiad, circumcircle
30.07.2019 15:42
Let $O'$ be center of $\odot{CDO}$ then $OO'$ is perpendicular bisector of $AB, CD$ the pretty trivial angle chase
30.08.2020 11:45
Let $O'$ be the center of $\odot{CDO}$ and $M$ be the midpoint of segment $AB$. We have $\angle APO' = \angle AMO' = 90^{\circ}$ and $\angle AQO' = \angle AMO' = 90^{\circ}$ so $AMQO'$ and $AMPO'$ are cyclic and we have $\angle AQM = \angle AO'M = \angle APM$. From that we get that $AMQP$ is cyclic too; as desired.
30.08.2020 15:12
Change $(CDO)$ in the statement on $\omega$ - circle through $C,D$ ($O$ is not needed). Let $M$ - midpoint of $AB$. If $O'$ - center of $\omega$ then $AO'$ - diameter of $APQ$. Since $O'$ is on perp bisector of $CD,AB$ we have $\angle AMO'=90^\circ$ and $M\in (APQ)$.
10.02.2022 04:10
Let $O_1$ be the center of $(COD)$ and note $(APQ)\cong (AO_1).$ We see if $X=\overline{AB}\cap(AO_1),$ $\angle AXO_1=90$ and so $X$ lies on the line perpendicular to $\overline{AB}$ that passes through $O_1.$ But $\overline{OO_1}$ is the perpendicular bisector of $\overline{AB}$ so $X$ is the midpoint of $\overline{AB}.$ $\square$
10.07.2022 10:55
I spent so much time on this Let $O'$ be the centre of $\odot(COD)$ and $M$ be the midpoint of $AB$, notice that $OO' \equiv O'M \perp AB$ and that $O' \in \odot(APQ)$ because $AP \perp PO', AQ \perp QO'$ which along with $AM \perp MO'$ gives us the desired conclusion $\blacksquare$