Question

In a right triangle ABC, a circle with a side AB as diameter is drawn to intersect the hypotenuse AC in P. Prove that the tangent to the circle at P bisects the side BC.​

Answer 1

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.