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M is mid point of AB
MP is perpendicular to AB
PM = PM (Common)
AM = BM (Construct)
^AMP = 1 rt. angle = ^BMP (Construct)
∇ APM = ∇ BPM (2 sides & incl. angle equal)
So
AP = BP
AP = BP
Alternative
Alternative
Consider triangle ABP with PA=PB and M mid point of AB.
Show
PM is perpendicular to AB
AP = PB (Construct)
AM = MB (Construct)
So
^PAM = ^PBM (∇ APM is isos.)
So
∇ APM = ∇ BPM (Two sides and inc. angle equal)
So
^PMA = ^PMB = 1 rt. angle (Since AMB is straight line)
^PMA = ^PMB = 1 rt. angle (Since AMB is straight line)
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