Prove that $A_1,B_1,C_1$ is collinear Let $ABC$ be a triangle inscribled inside circle $(O)$ . M is a point inside the triangle $ABC$ ($M \notin BC,CA,AB$)
$AM,BM,CM$ meets $(O)$ again at $A',B',C'$ respectively. Midperpendicular of $MA', MB', MC'$ meet $BC,CA,AB$ at $A_1,B_1,C_1$ respectively. 
1.Prove that $A_1,B_1,C_1$ is collinear
2.$M'$ is symmetric to M through $A_1B_1C_1$. Prove that $M'\in (O)$
Please give me some hint
Thanks
 A: Here's a vector approach.
With a view to invoking Menelaus' Theorem, we write
$$A_1 = a B + (1-a) C \qquad B_1 = b C + (1-b) A \qquad C_1 = c A + (1-c) B$$
and we hope to show
$$\frac{a}{1-a} \cdot \frac{b}{1-b} \cdot \frac{c}{1-c} = -1 \qquad (*)$$
Let $A^{\prime\prime} = M + \frac{1}{2}\frac{|A^\prime M|}{|AM|}(M-A)$ be the midpoint of $MA^\prime$. Since $\overrightarrow{A^{\prime\prime}A_1} \perp \overrightarrow{AM}$, we have
$$( A_1 - A^{\prime\prime} )\cdot( M - A ) = 0$$
so that
$$( \; a B + ( 1 - a ) C \; )\cdot( M - A ) = A^{\prime\prime}\cdot(M-A) = M\cdot(M-A) + \frac{1}{2}|AM||A^\prime M|$$
Thus,
$$a \; ( \; B - C \; )\cdot( M - A ) = ( M - C )\cdot(M-A) + \frac{1}{2}|AM||A^\prime M|$$
and
$$-(1-a) \; (\; B - C \; )\cdot( M - A ) = ( M - B )\cdot( M - A ) + \frac{1}{2}|AM||A^\prime M|$$
whence
$$\frac{a}{1-a} = - \frac{2\;\overrightarrow{CM}\cdot\overrightarrow{AM}+ |AM||A^\prime M|}{2\;\overrightarrow{AM}\cdot\overrightarrow{BM}+|AM||A^\prime M|}$$
Now, recall the Power of a Point theorem, which says that $M$ divides chords through it into sub-segments whose lengths have a constant product, denoted here as $p$:
$$|AM||A^\prime M| = |BM||B^\prime M| = |CM||C^\prime M| = p$$
This guarantees that the left-hand side of $(*)$ expands to a circular product:
$$
\left(- \frac{2\;\overrightarrow{CM}\cdot\overrightarrow{AM}+ p}{2\;\overrightarrow{AM}\cdot\overrightarrow{BM}+p}\right)
\left(- \frac{2\;\overrightarrow{AM}\cdot\overrightarrow{BM}+ p}{2\;\overrightarrow{BM}\cdot\overrightarrow{CM}+p}\right)
\left( - \frac{2\;\overrightarrow{BM}\cdot\overrightarrow{CM}+ p}{2\;\overrightarrow{CM}\cdot\overrightarrow{AM}+p}\right)
$$
which reduces to $-1$; according to Menelaus, $A_1$, $B_1$, $C_1$ are indeed collinear.
For now, I'll leave proof that $M^\prime$ lies on the circle as homework.
