In parallelogram $ABCD$, the bisectors of $\angle A$ and $\angle B$ meet at $P$. Then $\angle APB$ equals:
In parallelogram $ABCD$, the bisectors of $\angle A$ and $\angle B$ meet at $P$. Then $\angle APB$ equals:
$60^{\circ}$
$90^{\circ}$
$120^{\circ}$
Depends on the parallelogram
Solution
Adjacent angles of a parallelogram are supplementary, so $\angle A + \angle B = 180^{\circ}$. In $\triangle APB$: $\angle APB = 180 - (\angle A/2 + \angle B/2) = 180 - 90 = 90^{\circ}$.