Question

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

Answer 1

OB= OP 

so ∠ OBP= ∠ OPB.
∠ OPD is a right angle ( PD is a tangent).

Hence ∠ BPD= 90-∠OPB. 

Same as ∠ PBD= 90- ∠ OBP. 

as OBP and OPB ar equal, angle BPD= angle PBD and therefore DB=DP.

again.. angle OPX is a right angled  and ∠ APX= ∠DPC,

hence ∠ BPC= ∠BPD+∠ APX

as ∠ APX +∠APO =90 and also ∠APO +∠OPB=90,

⇒∠APX=∠OPB.
∴ ∠ BPC = ∠ BPD +∠ APX = ∠ BPD +∠ OPB = ∠OPD =90

BPC is therefor a right triangle and BC is a diameter. 

as DB=DP and D lies on the diameter BD would be equal to CD