Let $[ABC]$ be a acute-angled triangle and its circumscribed circle $\Gamma$. Let $D$ be the point on the line $AB$ such that $A$ is the midpoint of the segment $[DB]$ and $P$ is the point of intersection of $CD$ with $\Gamma$. Points $W$ and $L$ lie on the smaller arcs $\overarc{BC}$ and $\overarc{AB}$, respectively, and are such that $\overarc{BW} = \overarc{LA }= \overarc{AP}$. The $LC$ and $AW$ lines intersect at $Q$. Shows that $LQ = BQ$.
Problem
Source: 2019 Portugal p5
Tags: geometry, circumcircle, equal segments
W.R.O.N.G
11.10.2020 14:23
Short sketch: We claim that $Q$ is the midpoint of $AW$. First we show that $\Delta CBD$ and $\Delta CWA$ are similar. Obviously, $\angle AWC=\angle ABC$ and $\angle DCA=\angle PCA=\angle BCW$ and thus $\angle DCB=\angle ACW$. Also we see that $\angle WCQ=\angle BCA$ and since $CA$ is the median of $\Delta CBD$, we can conclude that $CQ$ is the median of $\Delta AWC$. Since $ALBW$ is an isosceles trapezoid with $Q$ as the midpoint of $AW$; $LQ=BQ$(This can be seen from symmetry).
iMqther_
29.06.2024 01:43
Firstly we have $AL=AP=BW$ because of the equal arcs so $LABW$ and $APBW$ are isosceles trapeziods thefore $AB//PW$. Now let $S$ be a point on $WP$ such that $AS//BW$ but we also have $AB//WS$ so $ASBW$ is a paralellogram $\implies$ $AB=AD=WS$. Now notice $AB //WP$ so $AD//WS$ which means $ADSW$ is a paralellogram $\implies$ $AW//DS$ then $AQ//DS$ so $\frac{BQ}{QS}=\frac{AB}{AD}=1$ so $Q$ is the midpoint of $BS$ $\implies$ $AQ=QW$. Notice that $AL=BW$, $AQ=QW$ and $\angle BWQ=\angle LAQ$ so we will have $\triangle LAQ \cong \triangle BWQ (S.A.S)$ $\implies$ $BQ=LQ$, as we wanted .
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