The bases of a trapezoid are equal to $a$ and $b$. It is known that a line can be drawn through the midpoint of one of its sides, dividing the trapezoid into two quadrilaterals, each of which can have a circle inscribed in it. Find the other lateral side of this trapezoid.
Let the extensions of sides $B M$ and $C N$ of the circumscribed quadrilateral $B C N M$ intersect at point $P$, and the extensions of sides $B C$ and $M N$ intersect at point $Q$. Prove that $P C+M Q=P M+C Q$.

## Solution

Let the line passing through the midpoint $M$ of the lateral side $A B$ of trapezoid $A B C D$ with bases $A D=a$ and $B C=b$, intersect the lateral side $C D$ at point $N$, and the extensions of the bases $A D$ and $B C$ at points $L$ and $Q$ respectively. Then $M Q=L M$ and $L A=B Q$.

Let the circle inscribed in the quadrilateral $B C N M$ touch its sides $B C, C N, N M$ and $B M$ at points $X, Y, Z$ and $T$. If the lines $A B$ and $C D$ intersect at point $P$, then

$$
\begin{aligned}
P C+M Q & =P Y-C Y+M Z+Q Z=P T-C X+M T+Q X= \\
& =(P T+M T)+(Q X-C X)=P M+C Q .
\end{aligned}
$$

Similarly, we can prove that

$$
P D+L M=L D+M P .
$$

Subtracting the first equality from the second and using the equalities $M Q=L M$ and $L A=B Q$, we get that

$$
\begin{gathered}
P D+L M-P C-M Q=L D+M P-P M-C Q \\
P D-P C+L M-M Q=L D+M P-P M-C Q \\
P D-P C=L D-C Q \\
C D=L D-C Q=(L A+A D)-(B Q-B C)=(L A-B Q)+(A D+B C)= \\
=A D+B C=a+b
\end{gathered}
$$

## Answer

$a+b$.