$A, B, C, D, E$ are five points on the same circle, so that $ABCDE$ is convex and we have $AB = BC$ and $CD = DE$. Suppose that the lines $(AD)$ and $(BE)$ intersect at $P$, and that the line $(BD)$ meets line $(CA)$ at $Q$ and line $(CE)$ at $T$. Prove that the triangle $PQT$ is isosceles. (I. Voronovich)
Problem
Source: 2013 France Training Test January p3 - original source : 2012 Belarus TST 1.2
Tags: geometry, isosceles, inscirbed, equal segments
blacksheep2003
24.09.2020 17:56
We have $B$ as the midpoint of arc $AC$ opposite $E$ and $D$ the midpoint of arc $CE$ opposite $A$ in $\triangle ACE$, so its incenter is $P$. We will show that $PQ = PT$. Inverting about $B$ with radius $PB$, we see that $Q$ maps to $D$, so $BQ \cdot BD = PB^2$, hence $\triangle PBQ$ is similar to $\triangle DBP$, therefore we obtain $\frac{PQ}{PD} = \frac{PB}{BD}$, or $PQ = \frac{PB \cdot PD}{BD}$. Similarly, inverting about $D$ with radius $PD$, we have $T$ maps to $B$, so $DT \cdot DB = PD^2$, hence $\triangle PDT$ is similar to $\triangle BDP$, therefore we obtain $\frac{PT}{PB} = \frac{PD}{BD}$, or $PT = \frac{PB \cdot PD}{BD} = PQ$. Thus, $PQ = PT$, so $\triangle PQT$ is isosceles, as desired.
WolfusA
24.09.2020 18:18
Solution by complex numbers is possible.
WolfusA
26.09.2020 20:09
Same problem as 2013 Italy EGMO TST p3 parmenides51 wrote: Let $ABCDE$ be a cyclic (non-entangled) pentagon where $AB = BC$ and $CD = DE$. Be $P$ the intersection between $AD$ and $BE$, let $Q$ be the intersection between $BD$ and $CA$, and let $R$ be the intersection between $BD$ and $CE$. (a) Prove that the triangle $PQR$ is isosceles. (b) Prove that the line $CP$ is perpendicular to the line $BD$. WolfusA wrote: Tricky, tricky, but let's start with $(b)$ and go to $(a)$. $\definecolor{A}{RGB}{0,90,255}\color{A}\fbox{Solution}$ First observe that $P$ is the incenter of $\triangle ACE$ so $CP$ is internal bisector of angle $BCD$. By trilium theorem $DP=CD\wedge BP=BC$ whence $\definecolor{A}{RGB}{0,255,0}\color{A}CP\perp BD$ and $\definecolor{A}{RGB}{255,0,0}\color{A}CP\perp RQ$. By inscribed angles $$\angle CRD=\angle BCD=\angle BQC\implies \definecolor{A}{RGB}{255,0,0}\color{A}CQ=CR.$$Thus$$\definecolor{A}{RGB}{255,0,0}\color{A}{PQ=PR}.\blacksquare$$#1783