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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.
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--- Europeon Mathematical Cup 2016: Junior cathegory question 2:
http://emc.mnm.hr/wp-content/uploads/2016/12/EMC_2016_Juniors_ENG.pdf

When I tried to solve this problem, I managed to prove that $$ E \in AB, $$ which reduces the question to proving that $$ CD \perp AE = AB \perp O_1O_2 \Rightarrow CD\parallel O_1O_2 \\ \text{ (with } O_1 \text{ and } O_2 \text{ the centers of } C_1 \text{ and } C_2 \text{ respectively).} $$

How can I do this?

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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

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  • $\begingroup$ 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. $\endgroup$ – Jonas De Schouwer Dec 16 '18 at 12:11

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