3
$\begingroup$

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 $a \perp AB$.

  • 1
    @Adam: Great. You've reduced the problem to the core issue, which isn't too tricky. Focus on $\triangle MP_1U_1$ and $\triangle MPV_1$.2012-05-08

1 Answers 1

4

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