Here is the problem:
In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find $MN$.
My question concerns part of the solution, which is this:
Extend ${CM}$ and ${CN}$ such that they intersect line ${AB}$ at points $P$ and $Q$, respectively. Since ${BM}$ is the angle bisector of angle B, and ${CM}$ is perpendicular to ${BM}$, so $BP=BC=120$, M is the midpoint of ${CP}$.
This isn't the full solution, but my question is why does $BC = BP$? This kind of makes sense because if the angle of B was larger, $AK$ and $AC$ would also be larger and $CM$ and $PM$ are also related, but could someone provide the proof?
You have two congruent triangles of $\triangle BMC$ and $\triangle BMP$. This is because $\angle BMC = \angle BMP = 90^{o}$, due to the requirement of $BM$ being perpendicular to $CM$, and $\angle CBM = \angle PBM$ as $BM$ is the angle bisector of $\angle CBA$. Thus, by the sum of angles in a triangle being $180^{o}$, the third angle in each triangle is also equal.
In addition, you have the common side $BM$, so obviously it's length is equal in both triangles. This is sufficient to prove triangle congruency. Thus, $BC = BP$ as they're the corresponding sides of $2$ congruent triangles.