Given a convex quadrilateral $ABCD$, side $AB$ is not parallel to side $CD$. The circle $O_1$ passing through $A$ and $B$ is tangent to side $CD$ at $P$. The circle $O_2$ passing through $C$ and $D$ is tangent to side $AB$ at $Q$. Circle $O_1$ and circle $O_2$ meet at $E$ and $F$. Prove that $EF$ bisects segment $PQ$ if and only if $BC\parallel AD$.
Problem
Source: China Mathematical Olympiad 1990 problem1
Tags: geometry unsolved, geometry
oneplusone
21.10.2013 07:49
Let $EF$ intersect $PQ$ at $M$. Extend $PQ$ to meet $O_1,O_2$ again at $X,Y$. Then $XM\cdot MP=ME\cdot MF=MQ\cdot MY$. Thus $MP=MQ\iff XQ=PY\iff AQ\cdot QB=XQ\cdot QP=QP\cdot PY=DP\cdot PC$. Extend $AB$ and $DC$ to intersect at $G$. Let $GB=a_1,GA=b_1,GC=a_2,GD=b_2$, and $a_1<b_1,a_2<b_2$. Now we have $GQ=\sqrt{a_2b_2}$ so $AQ\cdot QB=(b_1-\sqrt{a_2b_2})(\sqrt{a_2b_2}-a_1)$. Thus $AQ\cdot QB=DP\cdot PC\iff (a_1+b_1)\sqrt{a_2b_2}=(a_2+b_2)\sqrt{a_1b_1}\iff \sqrt{\frac{a_1}{b_1}}+\sqrt{\frac{b_1}{a_1}}=\sqrt{\frac{a_2}{b_2}}+\sqrt{\frac{b_2}{a_2}}\iff \frac{a_1}{b_1}=\frac{a_2}{b_2}\iff BC\parallel AD$ and we are done.
TheUltimate123
22.06.2021 08:05
Solved with Alan Bu, Alex Zhao, David Dong, Elliott Liu, Isaac Zhu, Jeffrey Chen, and Kevin Wu Define the linear function \(f:\mathbb R^2\to\mathbb R\) by \[f(\bullet)=\operatorname{Pow}(\bullet,(APB))-\operatorname{Pow}(\bullet,(CQD)).\]Let \(X=\overline{AB}\cap\overline{CD}\), so \(XP^2=XA\cdot XB\) and \(XQ^2=XC\cdot XD\). We can find that \begin{align*} QA\cdot QB&=(XQ-XA)(XB-XQ)\\ &=(XA+XB)\sqrt{XC\cdot XD}-XA\cdot XB-XC\cdot XD. \end{align*}It follows that \begin{align*} \overline{EF}\text{ bisects }\overline{PQ}\iff \operatorname{Pow}(Q,(APB))&=\operatorname{Pow}(P,(CQD))\\\iff QA\cdot QB&=PC\cdot PD\\\iff (XA+XB)\sqrt{XC\cdot XD}&=(XC+XD)\sqrt{XA\cdot XB}\\\iff \sqrt{\frac{XA}{XB}}+\sqrt{\frac{XB}{XA}}&=\sqrt{\frac{XC}{XD}}+\sqrt{\frac{XD}{XC}}\\\iff \frac{XA}{XB}&=\frac{XD}{XC}\iff \overline{BC}\parallel\overline{AD}, \end{align*}as desired.