A pentagon $ABCDE$ is such that $ABCD$ is cyclic, $BE\parallel CD$, and $DB=DE$. Let us fix the points $B,C,D,E$ and vary $A$ on the circumcircle of $BCD$. Let $P=AC\cap BE$, and $Q=BC\cap DE$. Prove that the second intersection of circles $(ABE)$ and $(PQE)$ lie on a fixed circle.
Problem
Source: Own. IMO 2022 Malaysian Training Camp 1
Tags: geometry
P2nisic
06.09.2023 16:20
Lemma:if we have an isosceles trapezoid $ABCD$ and $E$ a point on $CD$.Also $Z$ on $AE$ such that $BZ//AC$ and $H=BZ\cap AD$ then: $\frac{ZH}{HB}=\frac{DE}{EC}$ Proof: $\frac{ZH}{HB}=\frac{AZ}{AB}\frac{sin\angle ZAH}{sin\angle HAB}=\frac{sin\angle ABZ}{sin\angle AZB}\frac{sin\angle ZAH}{sin\angle HAB}=\frac{sin\angle ACD}{sin\angle EAC}\frac{sin\angle DAE}{sin\angle ADC}=\frac{sin\angle ACD}{sin\angle ADC}\frac{sin\angle DAE}{sin\angle EAC}=\frac{AD}{AC}\frac{sin\angle DAE}{sin\angle EAC}=\frac{DE}{EC}$ Now back to the problem: Let $Z=EQ\cap AC,Y=EB\cap (BCD)$ we can see that: $\angle CYB=\angle YBD=\angle BED\Rightarrow YC//DE$ and $\angle ZEB=\angle CYB=\angle ZAB\Rightarrow Z\epsilon (ABE)$ We are going to prove that $E,X,D,Y$ are cyclic by the coaxality lemma we need to prove that:$\frac{Pow(Y,(ABE)))}{Pow(Y,(EPQ))}=\frac{Pow(D,(ABE)))}{Pow(D,(EPQ))}$ but this leads to: $\frac{Pow(Y,(ABE)))}{Pow(Y,(EPQ))}=\frac{Pow(D,(ABE)))}{Pow(D,(EPQ))}\Leftrightarrow \frac{YE*YB}{YE*YP}=\frac{DE*DZ}{DE*DQ}\Leftrightarrow \frac{YB}{YP}=\frac{DZ}{DQ}\Leftrightarrow \frac{BP}{YP}=\frac{QZ}{DQ}$ Which is true by the lemma
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