Chords Problem?

Question

Chords AC and BD are perpendicular to each other and intersect at point G. In triangle AGD the altitude from G meets AD at E, and when extended meets BC at P. Prove BP = PC.

A little help please?

Answer 1

Angle DAC = Angle DBC = x, say. (Angles subtended by the same arc DC at two different points on the circle)
Similarly, Angle ADB = Angle ACB = 90 - x (since the other angles in the triangles formed are 90, x in each)
Now angle EGD = x (since angle GED is 90 and the other is 90 - x)
=> Angle BGP = x (vertically opposite)
Thus triangle BPG is isosceles (two angles are equal to x)
Thus GP = PB - (i)
Similarly triangle CPG is isosceles, equal angles being equal to 90-x. Thus GP = PC - (ii)
From (i) and (ii) PB = PC or to say the line GP is a median.