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 Apr 22, 2019
edited by Guest  Apr 25, 2019
 #1
avatar+104962 
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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

 

 

cool cool cool

 Apr 22, 2019

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