Let $M, H$ and $N$ be the feet of the perpendiculars dropped from point $K$ to $AC, AB$ and $BC$ respectively. Points $M$ and $H$ lie on the circle with diameter $AK$. From the theorem about the angle between a tangent and a chord, $\angle K M H = \angle K A H = \angle K A B = \angle K B N$.

Points $N$ and $H$ lie on the circle with diameter $BK$, so $\angle K H N = \angle K B N = \angle K M H$.

Similarly, $\angle K H M = \angle K N H$, hence triangles $K M H$ and $K H N$ are similar by two angles. Therefore, $K H : K N = K M : K H$, from which

$K H^{2} = K N \cdot K M = 144$

## Answer

12.