In triangle $ABC$ bisectors $AA_1$ and $BB_1$, are drawn. prove that the distance away from any point $M$ to $A_1B_1$ In triangle $ABC$ bisectors $AA_1$ and $BB_1$,are drawn. Prove that the distance away from any point $M$ on $A_1B_1$ to line $AB$ is equal to the sum of the distances from $M$ to $AC$ and $BC$
I got this problem off Viktor Prasolov's Plane Geometry and I've been on it for hours and still nothing.
 A: Outline of algebraic proof: The statement is true for $M = A_1$ or $M=B_1$. Let's rephrase the problem as 

The distance between $M$ and $AC$, plus the distance between $M$ and $BC$, minus the distance from $M$ to $AB$, equals $0$.

Let $M = t\cdot A_1 + (1-t)\cdot B_1$ (think like with vectors here if you know them) and show that the quantity defined in the above rephrasing, which should always equal zero, is a first-degree expression in $t$. Since it's zero twice, it's zero for all $t$ (at least between $0$ and $1$).
A: For simplicity, I re-phrase the question as:-

AP and BQ are the angle bisectors of $\triangle ABC$. M is a point on PQ. $MM_c$, $MM_a$, and $MM_b$ are perpendiculars to the sides AB, BC and CA respectively. Prove that $M_c = M_a + M_b$.


Construction:-
(1) Through P, draw perpendiculars $P_b$, $P_c$ to AC and AB respectively. Note that $PP_b = PP_c$.
(2) Through Q, draw perpendiculars $QQ_a$ and $QQ_c$ to BC and AB respectively.
(3) Through Q, draw line $\lambda$ parallel to AB cutting $MM_c$, and $PP_c$ at M’ and P’ respectively. Note that if $\lambda$ is at a distance h from AB, then $P’P_c = M’M_c = QQ_c = QQ_a = h$.
Let $QM : MP = 1 : r$.
Note that the grey-shaded triangle contains two similar triangles. The same is true for the blue and green triangles. Therefore, we have:-
$\dfrac {MM’}{PP’} = \dfrac {1}{1 + r}$......(1)
$\dfrac {MM_b}{PP_b} = \dfrac {1}{1 + r}$……(2)
$\dfrac {MM_a}{QQ_a} = \dfrac {r}{1 + r}$. That is, $MM_a = \dfrac {rh}{1 + r}$......(3)
From (1), $MM_c = MM’ + h = \dfrac {PP’}{1 + r} + h = \dfrac {PP’ + h + rh}{1 + r}$
$= \dfrac {PP_c + rh}{1 + r} = \dfrac {PP_b + rh}{1 + r}$
$= \dfrac {MM_b (1 + r) + rh}{1 + r}$ [from (2)]
$= MM_b + \dfrac {rh}{1 + r} = MM_b + MM_a$ [from (3)]
