Find the distance between the points of tangency of the circles inscribed in triangles $A B C$ and $C D A$ with side $AC$, if

a) $A B=5, B C=7, C D=D A$;

b) $A B=7, B C=C D, D A=9$.
The distance from the vertex of a triangle to the nearest point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side of the triangle ($x = p - a$).

## Solution

a) Let the inscribed circle of triangle $ABC$ touch side $AC$ at point $K$, and the inscribed circle of triangle $CDA$ touch at point $M$. Since the distance from the vertex of a triangle to the point of tangency with the inscribed circle is equal to the difference between the semiperimeter and the opposite side of the triangle, then

$$
A K = \frac{A B + A C - B C}{2}, A M = \frac{A C + A D - D C}{2}
$$

Then

$$
K M = |A K - A M| = \frac{|A B + D C - B C - A D|}{2} = |7 - 9| = 1.
$$

## Answer

a) 1; b) 1.