22. In a right triangle ABC a circle is drawn with AB as the diameter which interscet
hypotenuse AC at point P. Prove PB = PC.
Answers
Answered by
2
The required proof would be to prove that points B,P and D lie on a semicircle with centre D.
Explanation: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.
Explanation: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.
Similar questions