Let $P$ be the base of the perpendicular dropped from vertex $C$ of the smaller base $BC$ of the isosceles trapezoid $ABCD$ to its larger base $AD$. Find $DP$ and $AP$, if the bases of the trapezoid are equal to $a$ and $b (a > b)$.
Let $Q$ be the base of the perpendicular dropped from vertex $B$ to $A D$. From the equality of right triangles $A B Q$ and $D C P$, it follows that $A Q=D P$, and since $B C P Q$ is a rectangle, $P Q=B C=b$. Therefore, $D P=1 / 2(A D-P Q)=1 / 2(a-b), A P=A D-D P=a-1 / 2(a-b)=1 / 2(a+b)$.

## Answer

$1 / 2(a-b), 1 / 2(a+b)$