Prove it's perpendicular

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Raiyan Jamil
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Prove it's perpendicular

Unread post by Raiyan Jamil » Mon Aug 15, 2016 8:40 pm

Let one of the intersections of two circles with centres $O1$ and $O2$ be $P$. A common tangent touches the circles at $A$,$B$ respectively.Let perpendicular from $A$ to $BP$ meet $O1O2$ at $C$. Prove that $AP$ is perpendicular to $PC$.
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tanmoy
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Re: Prove it's perpendicular

Unread post by tanmoy » Mon Aug 15, 2016 10:13 pm

The main idea of this solution is:If $\measuredangle APC=\dfrac {\pi} {2}$,then $CP$ must pass through the antipode of $A$,say,$A_{1}$ in $(O_{1})$.

Let $(O_{1}) \cap (O_{2})={P,Q}$.Then it is well known that $PQ$ passes through the midpoint of $AB$,say,$M$.

Now,let $O_{1}O_{2} \cap MQ=R$,then $\measuredangle O_{2}BM=\measuredangle MQO_{2}=\dfrac {\pi} {2}$.So,$O_{2}BMQ$ is cyclic.So,$\measuredangle BMP=\measuredangle BMQ=\measuredangle AO_{1}C$ and $\measuredangle O_AC=\dfrac {\pi} {2}-\measuredangle BAC=\measuredangle ABP$.So,$\triangle AO_{1}C \sim \triangle BMP$.So,$\triangle AA_{1}C \sim \triangle BAP$.So,$\measuredangle BAP=\measuredangle AA_{1}P=\measuredangle AA_{1}C$.Therefore, $A_{1},C,P$ are collinear.So,done.
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