In and are the midpoints of and respectively. Lines and intersect at . If , prove that .

**Solution**

Connect points Q and R. Then by midpoint theorem, and .

Let M be the intersection of SB and QR.

Since SB is the height of , then MB is the height of (i.e. ). It follows that M is the midpoint of SB and by SAS postulate. Hence, . As R is the midpoint of CB, then , and so . Also, .

Due to the midpoint theorem, PQRC is a parallelogram, which implies that .

In trapezoid PQRS, opposite angles and add up to 180. Therefore PQRS is a cyclic quadrilateral.

Now (angle subtended by arc PQ). , then . Then . Therefore, .

*Addendum:*

Clearly, a cyclic trapezoid would be formed also by connecting the midpoint of each side and the perpendicular foot on another side.

Theorem:The quadrilateral formed by the midpoint of each side and a perpendicular foot of a triangle is a cyclic trapezoid.