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Guest Apr 22, 2019

edited by
Guest
Apr 25, 2019

#1**+1 **

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

CPhill Apr 22, 2019