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See the following :
Connect the center of the circle O with M....since M is the midpoint of chord BC, then the diameter intersects this chord at right angles....so angle CMF = angle CMP = 90°
Note that we have two triangles, CPM and BPM
And the diameter bisects chord BC...so....CM = BM
And PM is common
So...by Leg-Leg Congruency, triangle CPM is congruent to Triangle BPM
And angle CPM = angle EPD [ vertical angles]
And angle BPM = angle EPA [ vertical angles ]
But angle CPM = angle BPM.....so angle CPM = angle EPA
And inscribed angles DCB and DAB intersecpt the same arc....so they are equal
So....by AA congruency.....triangle CPM is similar to triangle APE
And angle CMP = 90° = angle AEP
But MP is perpendicular to BC
So MP produced must also be perpendicular to AD