In $\Delta ABC$, $AD \perp BC$ at $D$. $E,F$ lie on line $AB$, such that $BD=BE=BF$. Let $I,J$ be the incenter and $A$-excenter. Prove that there exist two points $P,Q$ on the circumcircle of $\Delta ABC$ , such that $PB=QC$, and $\Delta PEI \sim \Delta QFJ$ .
Problem
Source: 2019 China TST Test 3 P5
Tags: geometry, incenter, circumcircle
29.03.2019 18:08
Let point $P$ and $Q$ are in circumcircle of $\Delta ABC$, and $\angle API=90$, and $PQ//BC$. Let the midpoint of segment $PQ$ is $N$, and midpoint of small arc $BC$ is $M$. We will show that $P$ and $Q$ are satisfy the condition. We are enough to show that $\Delta EIP\sim \Delta BMN$.(Because $B$ is midpoint of $EF$ and $M$ is midpoint of $IJ$. <Step 1>: $\angle EIP=\angle BMN$
<Step 2> $\Delta EIP\sim \Delta BMN$
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30.03.2019 04:39
16.07.2019 06:14
IN,JN' perp BC, (AI)-(O)=P, AI cuts BC and (O) at L,K, AO-(O)=R then D(ALIJ)=-1 so BDI=FDJ so BEI=BFJ so EI, BR, FJ collinear. From KJ^2=KN'.KQ so AQJ=AQK+KQJ=ACK+KJN=ALB+LIN=90 so Q,R,J collinear. Because RI=RJ and SR bisector so SIRJ cyclic so PIE=QJF. By Mene AJ/AI=FJ/EI (SE=SF) but API~AQJ so QJ/JF=PI/IE so PIE~QIF hence we have the conclusion
13.01.2021 14:32
on the intersection of E,F and I,J and P,Q we get those triangles to be congruent