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

Answer 2

Answer:

Call the circle’s center O; the tangent intersects BC at D. Extend radius OP to point Z at an indefinite distance. The aim is to show that CD = DB = DP.

Angle OPA equals angle OAP, so angle OAP equals angle CPZ. Angle CPZ plus angle CPD equals 90 degrees, and angle BCA plus angle CAB equals 90 degrees, so angle BCA equals angle CPD. So triangle CPD is isosceles — length of CD equals length of PD, which equals length of BD.

Step-by-step explanation:

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