Question

Given a right-angled triangle ABC in which ∠B = 90°, a circle is drawn with AB as diameter intersecting the hypotenuse AC at P. Prove that tangent to the circle at P bisects BC.

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Answer 1

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.

Thus, the tangent to the circle at P bisects BC.