Consider two circles $O_1$ and $O_2$, the intersection of $O_1$ and $O_2$ are $A$ and $B$.
Let $M$ be the point on $O_1$, $N$ on $O_2$, $M,N$ moving clockwise on $O_1,O_2$ and $\angle{AO_1M }=\angle{AO_2N }$.
Show that the midperpendiculars of $MN$ pass through $P$ is a constant point.
I tried circles of Apollonius but failed. Help me,pls.
This has notihng to do with Apollonius.
Step 1: Let line $O_1M$ meet line $O_2N$ at $Q$. Then since $$\angle QO_1A = \angle QO_2A $$ we see that points $O_1, O_2, Q$ and $A$ are concyclic and so $\angle O_1QO_2 = \angle O_1AO_2$ is constatnt.
Step 2: Let $S$ be a midpoint for segment $O_1O_2$. Let us prove that $SP$ is constant, that is, $P$ is on circle with center at $S$. We use (position) vectors:
\begin{eqnarray} 4\cdot SP^2 &=& 4\vec{SP}^2\\ &=& 4(P-S)^2\\ &=& 4 \Big({1\over 2}(M+N)-{1\over 2}(O_1+O_2)\Big)^2\\ &=& 4 {1\over 4}\Big((M-O_1)+(N-O_2)\Big)^2\\ &=& (\vec{O_1M}+\vec{O_2N})^2\\ &=& \vec{O_1M}^2+\vec{O_2N}^2 +2\vec{O_1M}\cdot \vec{O_2N}\\ &=& r_1^2+r_2^2+2r_1r_2\cos \angle(\vec{O_1M}, \vec{O_2N})\\ &=& constant \end{eqnarray} Here we use step 1 conslusion that $\angle (\vec{O_1M}, \vec{O_2N}) = \angle O_1QO_2$.
Step 3: Try to finish yourself... :)