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I need help with the following problem.

Given three circles $k, k_1, k_2$. $k_1$ and $k_2$ touch internally $k$ at points $M$ and $N$ respectively. $a$ is the common interior tangent to $k_1$and $k_2$ at points $R$ and $S$. $MR \cap k = A$ and $NS \cap k = B$. Prove that $a \perp AB$.

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What part of it do you need help with? – Théophile May 8 '12 at 19:05
I don't have any idea how to prove it. Maybe you can give me a hint? – Adam May 8 '12 at 19:10
I need clarification - aren't $M$ and $N$ points on $k$? If so, doesn't this imply $M=A$ and $N=B$? – process91 May 8 '12 at 19:14
HINT: First concentrate on one internal circle, $k_1$. Say $P$ and $P_1$ are the centers of $k$ and $k_1$, respectively. For point $U_1 \neq M$ any (other) point on $k_1$, let $V_1$ be the "second intersection point" of $MU_1$ with $k$. What can you say about $PV_1$ and $P_1U_1$? ... ... When $U_1$ coincides with your point $R$ (so $V_1$ coincides with $A$), what can you say about $PA$? ... ... How does a similar understanding of $k_2$ and $PB$ help? (Pay attention to how $P_1R$ and $P_2S$ relate.) – Blue May 8 '12 at 21:40
@Adam: Great. You've reduced the problem to the core issue, which isn't too tricky. Focus on $\triangle MP_1U_1$ and $\triangle MPV_1$. – Blue May 8 '12 at 22:40
up vote 4 down vote accepted

enter image description here

Here is the diagram of the construction described, with the addition of the centers of $k$, $k_1$, and $k_2$; $T$, $P$, and $Q$.

Note that $\triangle PMR\simeq\triangle TMA$ and $\overline{PR}||\overline{TA}$.

Note that $\triangle QNS\simeq\triangle TNB$ and $\overline{QS}||\overline{TB}$.

Since $\overline{PR}\perp\overline{RS}$, we have $\overline{TA}\perp\overline{RS}$.

Since $\overline{QS}\perp\overline{RS}$, we have $\overline{TB}\perp\overline{RS}$.

Therefore, $\overline{TA}||\overline{TB}||\overline{AB}$.

Thus, $\overline{AB}\perp\overline{RS}=a$.

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