# JEMC 2016/2: Two circles C1 and C2 intersect at points A and B. Let P, Q be points on circles C1, C2 respectively, such that |AP| = |AQ|.

Two circles C1 and C2 intersect at points A and B. Let P, Q be points on circles C1, C2 respectively, such that |AP| = |AQ|. The segment P Q intersects circles C1 and C2 in points M, N respectively. Let C be the center of the arc BP of C1 which does not contain point A and let D be the center of arc BQ of C2 which does not contain point A. Let E be the intersection of CM and DN. Prove that AE is perpendicular to CD.

--- Europeon Mathematical Cup 2016: Junior cathegory question 2:
When I tried to solve this problem, I managed to prove that $$E \in AB,$$ which reduces the question to proving that $$CD \perp AE = AB \perp O_1O_2 \Rightarrow CD\parallel O_1O_2 \\ \text{ (with } O_1 \text{ and } O_2 \text{ the centers of } C_1 \text{ and } C_2 \text{ respectively).}$$
We can solve that $$E$$ is the orthocenter of $$\triangle ACD$$. In the picture above, $$\angle BAD =\frac{\angle BAQ}2, \angle BAC = \frac{\angle BAP}2.$$ So $$\angle CAD = \frac12 \angle PAQ$$. Moreover, $$\angle MCA = \angle QPA = \angle PQA = \frac{180^\circ - \angle PAQ}2.$$ So $$\angle MCA + \angle CAD = 90^\circ$$, or $$CM\perp AD$$. Similarly $$DN\perp AC$$.
Note: it's not immediate for me to see that $$A,E,B$$ are colinear. Would be nice if you can include the proof in your question.s
• I assumed that $N \in AC$ and $M \in AD$ (because it was like this in my draft drawing). In that case, it's easy to see that (with angles modulo 180°) $NCM = ACM = APM = APQ = AQP = AQN = ADN = MDN$, from which follows that $MNCD$ is cyclic. Hence, $E$ must lie on the radical axis of $C1$ and $C2$ ($AB$). Unfortunately, it isn't always the case that $N \in AC$ and $M \in AD$, so my proof is actually worthless (even though I'm sure there must be a way to prove that $MNCD$ is cyclic). But I'll leave my question as it is, as I'm still interested in how the lemma at the end can be proven. – Jonas De Schouwer Dec 16 '18 at 12:11