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

 

 

 

OP will be perpendicular to tangent hence angle OPD=90°

 

In triangle OAP OA=OP =radius

Therefor, angle OAP=Angle OPA let it be ?.

Angle DPC=180-(AngleOPD +Angle OPA)=180-90-?=90-?

In right angled triAngle ABC Angle C=180-(Angle A+ Angle B)=180-90-?=90-?

 

In triAngle PDC Angle P=Angle C= 90-?

therefor DP=DC

Now DP and DC are the tangents to same circle from the point D implies that DP=DB and we have proved that DP=DC

Using both we can say DB=DC hence D id mid point of BC.

 

Hence proved that tangent at P bisects BC.