Help with these two hard problems:
Let M be a point inside the angle A. From point M perpendiculars MP and MQ are dropped to the angle sides. From point A, perpendicular AK is dropped to segment PQ. Prove that angle PAK = angle MAQ.
Let ABC be a triangle and angle B = 60. Angle bisectors AD and CE meet at the point O. Prove that OD = OE.
... and here's the answer for number 13:
Then, (angle sum of )
Now is an exterior angle of = sum of interior opposite angles
is a cyclic quadrilateral (exterior angle = interior opposite angle )
Now the angle bisectors of a triangle are concurrent is the bisector of
(angles in same segment)