MathDB

I have a very good solution of this but I want to see others.
Let the midpoint

M

of the side

AB

of an inscribed quardiletar,

ABCD

.Let

P

the point of intersection of

MC

with

BD

. Let the parallel from the point

C

to the

AP

which intersects the

BD

S

. If

CAD

angle=

PAB

angle=

\frac{BMC}{2}

angle, prove that

BP=SD

Problem Statement