The Cunning Midpoints

We consider a general case.
Let $BC = a, CA = b, AB = c$ in $\triangle ABC$ .
Also, let $b \geq c$ so that the configuration is as shown in the figure.

As $AD$ is the angle bisector of $A$ , it divides the base BC in the ratio $\dfrac{BD}{DC} = \dfrac{c}{b}$
This gives us $BD = \dfrac{ac}{b + c}$ .

$A_1$ is the midpoint of $BC$ , so $BA_1 = \dfrac{a}{2}$ .
$D A_1 = BA_1 - BD = \dfrac{a}{2} - \dfrac{ac}{b + c}$

Now, $A_1 B_1$ is parallel to $AB$ by the midpoint theorem.
$\Rightarrow \triangle A_1 D K$ and $\triangle BDA$ are similar.

Thus, $\dfrac{A_1 K}{A_1 D} = \dfrac{c}{BD}$ .
$\Rightarrow A_1 K = \dfrac{b + c}{a}\cdot\left( \dfrac{a}{2} - \dfrac{ac}{b + c}\right) = \dfrac{b + c}{2} - c = \dfrac{b - c}{2}$ .

$A_1 K = \dfrac{|{b - c}|}{2}$ in general.

Here, $b = 13, c = 5 \Rightarrow A_1 K = 4$

We see that $A_1 K$ was independent of $a$ !

Here is a non-bash solution:

Let $\angle BAC=2\theta$ , so that we have $\angle BAK=\theta=\angle KAB_1$ .

By midpoint theorem, we have $A_1B_1 \| AB$ , so that $\angle KB_1C=\angle BAB_1=2\theta$ , that is, $\angle AB_1K=\pi-2\theta$ .

In $\Delta AB_1K$ , we have $\angle KAB_1=\theta, \angle AB_1K=\pi-2\theta$ , so that $\angle AKB_1=\theta$ , so we have $\Delta AB_1K$ is isosceles,

with $AB_1=B_1K$ ,

and $A_1K=B_1K-B_1A_1=AB_1-B_1A_1=\frac{AC}{2}-\frac{AB}{2}$

(since $B_1$ is the midpoint of $AC$ and by midpoint theorem we have $B_1A_1=\frac{BA}{2}$ )

Even bashing helps .

Let $A=(0,5) ; B=(0,0) ; C=(12,0)$ . Then we have

$A_1=(6,0); B_1=(6,2.5) ; K=(6,-4)$ .

$K$ is finded out by using method as follows,

Equation of angle bisector of ang $BAC$ = $3x +2y-10$

Thus solving it with $x=6$ , we have desired result. Thus $A_1 K=4$