# Find the angle made by the intersection point and midpoints of the diagonals of a cyclic quadrilateral

In cyclic quadrilateral $ABCD$, let $AD$ and $CB$ meet at $P$. Let $E$ and $F$ be the midpoints of $DB$ and $CA$ respectively. Find $\angle PEF$ in terms of $\alpha, \beta, \gamma, \delta$, which are the angles of the cyclic quadrilateral.

• If the circle rotates, then the angles at $A,B,C,D$ stay constant. but it seems the angle $PEF$ will vary. – coffeemath Jul 26 '17 at 7:13
• @coffeemath what do you mean by "If the circle rotates"? It seems that once you have a cyclic quadrilateral, all of the above is fixed? – Plato Jul 26 '17 at 7:22
• Plato-- I guess my comment is not right, but maybe I was trying to express the ambiguity in definition of angle PEF noted in Artino's example below. – coffeemath Jul 26 '17 at 16:25

Angle $\angle PEF$, in general, does not depend only on the angles of the cyclic quadrilateral. See below an example, with two cyclic quadrilaterals having the same angles (they have parallel sides) but different values for $\angle PEF$. You also need to know the radius of the circle.