Given a cyclic quadriteral $ABCD$. The circumcenter lies in the quadriteral $ABCD$. Diagonals $AC$ and $BD$ intersects at $S$. Points $P$ and $Q$ are the midpoints of $AD$ and $BC$. Let $p$ be a line perpendicular to $AC$ through $P$, $q$ perpendicular line to $BD$ through $Q$ and $s$ perpendicular to $CD$ through $S$. Prove that $p,q,s$ intersects at one point.
Problem
Source: Polish MO 2022 P2
Tags: geometry, circumcircle, perpendicular
Steve12345
13.02.2022 01:56
Let $p,q$ cut $BD,AC$ again at $E,F$. It is enough to prove $EF \parallel CD$ by orthocenter $\Leftrightarrow \frac{SC}{SD} = \frac{CF}{DE}$ but $\frac{CF}{DE} = \frac{CQ}{DP} = \frac{BC}{AD} = \frac{SC}{SD} \ \blacksquare.$
MathsLion
13.02.2022 17:49
Here's a slightly different solution. Let $p \cap q=K$. Points $E, F$ are the "same" points (analogously defined) in similar triangles $ADS$ and $BCS$. That gives us $\frac{AF}{FS}=\frac{BE}{ES}$ which is equivalent to $EF \parallel AB$. Now angle chase showed in the figure finishes.
Attachments:
timon92
13.02.2022 21:19
This problem was proposed by Burii.
Mahdi_Mashayekhi
31.03.2022 07:58
Let $p$ meet $AC$ at $K$ and $q$ meet $BD$ at $T$ and $p,q$ meet at $X$ and $XS$ meet $DC$ at $H$. we want to prove $\angle XHD = \angle 90$. we will prove $XTHD$ is cyclic. $\angle TDH = \angle BAS$ and $TXH = \angle TKS$ so we need to prove $KT || AB$. Note that $ASD$ and $BSC$ are similar and $P,Q$ are midpoint's of $AD,BC$ so $K,T$ same points in two similar triangles so $\frac{AK}{KS} = \frac{BT}{TS} \implies KT || AB$. we're Done.