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.
Asked by yashjain | 28th Nov, 2012, 04:18: PM
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-?
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.
Answered by | 29th Nov, 2012, 10:52: PM
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