Prove that $AC_1+AC=BC_1+BC$

Question

In $ABC$ triangle let $w$ be the inscribed circle inside $ABC$ and let $MM_1$ be a diameter of $w$.

If $CM_1$ and $AB$ intersect at $C_1$, Prove that $AC_1+AC=BC_1+BC$.

MY ATTEMPT:

I haven't achieved much. I can tell that if we prove $AC_1=BM$, then the problem would be solved. (A solution using homothety would be nice)

Answer 1

A very standard problem.

Notice that $A'B'$ is parallel to $AB$ with tangent point at $M_1$, this means if we amplify the whole $\triangle CA'B'$ to $\triangle CAB$, the outer circle will touch at the extension of $CM_1$, which is $C_1$.

Now since $AM_2+AM_4=BM_3+BM_5$, we have $AC_1+AM=BM+BC_1$, which implies $AC_1=BM$ and you are good.