In a right angled triangle ABC a circle is drawn with AB as a diameter which intersect hypotenuse AC at point P. Prove PB = PC
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In triangle OAP
OA=OP =radius
Therefore,
angle OAP=Angle OPA = x
Angle DPC= 180 - (Angle OPD +Angle OPA)
=180 - 90 - x
=90-x
In right angled triangle ABC
Angle C=180 - (Angle A+ Angle B)
=180-90-x
=90-x In triangle PDC
Angle P=Angle C= 90-xtherefore
DP=DC
As DP and DC are the tangents of same circle from the point D so in a same way DP=DB
and thus DP=DC
or DB=DC
hence D is mid point of BC.
Hence proved that tangent at P bisects BC
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