In $\Delta ABC$, $AB = 14, BC = 16, AC = 26$. $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$.

Question

In $\Delta ABC$, $AB = 14, BC = 16, AC = 26$. $M$ is the midpoint of $BC$ and $D$ is the point on $BC$ such that $AD$ bisects $\angle BAC$. Let $P$ be the foot of the perpendicular from $B$ to $AD$ . Find $PM$ .

What I Tried: Here is a picture :-

The first thing which I noticed is that, $AC \parallel PM$ from Geogebra. I tried everything to prove that from the diagram but without success. I got no idea as I don't think checking for areas will help, and I tried angle-chasing to the most. And obviously, I am in $50$ steps away from solving the problem.

Can someone show me how $AC \parallel PM$ ? Thank You.

Answer 1

$P$ is the midpoint of $BK$. $M$ is the midpoint of $BC$.

So $\triangle BPM \sim \triangle BKC$ (by A-S-S).

$KC = AC - AK = 26 - 14 = 12$

So $PM = 6$.

Answer 2

Hint:

PM||AC because it connects midpoints of BC and BK.P is mid point of BK, do you know why?

Answer 3

All the side length $a,b,c$ are known, hence, we can get all the angles $\alpha,\ \beta,\ \gamma$;

$|BM|=\tfrac12\,a$,

$|BP|$ can be found from $\triangle ABP$,

$\angle PBM=\tfrac12\,\alpha+\beta-90^\circ$,

use the cosine rule for $\triangle BMP$ to find $|MP|$:

\begin{align} |MP|^2=|BM|^2+|BP|^2-2\cdot|BM|\cdot|BP|\cdot\cos\angle PBM . \end{align}