Question

In a right angle 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

Answer:

OB= OP , radii of the given circle. Hence angle OBP= angle OPB.

angle OPD is a rt angle ( PD is a tangent). Hence angle BPD= 90-OPB. Similarly, angle PBD= 90- angle OBP. Since OBP and OPB ar equal, angle BPD= angle PBD and therefore DB=DP.

Next, angle OPX is rt. angle and angle APX= angleDPC, hence angle BPC= Angle BPD+angle APX

Since, angle APX +APO =90 and also APO +OPB=90,this means APX=OPB.

Hence angle BPC = Angle BPD +angle APX = Angle BPD +angle OPB = angle OPD =90

BPC is therefor a right triangle and BC is a diameter. Since DB=DP, and D lies on the diameter, BD would be equal to CD.

Answer 2

Step-by-step explanation:

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