3a On the side of a triangle, four points $K, P, H$, and $M$ are taken, which are respectively the midpoint of this side, the foot of the bisector of the opposite angle of the triangle, the point of tangency with this side of the circle inscribed in the triangle, and the foot of the corresponding altitude. Find $K H$, if $K P=a, K M=b$.

#
Let the specified points lie on side $AB$ of triangle $ABC$, where $AB=x$, $BC=y$, $AC=z$, and for definiteness, we will assume that $y>x$. Then

$$
B K=\frac{1}{2} x, B P=\frac{x y}{y+z}, B H=\frac{1}{2}(x+y-z)
$$

$$
B M=y \cos \angle B=\frac{y\left(x^{2}+y^{2}-z^{2}\right)}{2 x y}=\frac{x^{2}+y^{2}-z^{2}}{2 x}
$$

Next, we find:

$$
\begin{gathered}
K P=B P-B K=\frac{x(y-z)}{2(y+z)}, K H=B H-B K=\frac{1}{2}(y-z) \\
K M=B M-B K=\frac{y^{2}-z^{2}}{2 x}
\end{gathered}
$$

Thus,

$$
K P \cdot K M=\frac{x(y-z)}{2(y+z)} \cdot \frac{y^{2}-z^{2}}{2 x}=\frac{1}{4}(y-z)^{2}=K H^{2}
$$

Therefore, $K H=\sqrt{K P \cdot K M}=\sqrt{a b}$.

## Answer

$\sqrt{a b}$.