In a right triangle ABC in which angle B = 90', a circle is drawn with AB as diameter intersecting the hypotenuse ACat P. Prove that the tangent to the circle at P visects BC.

Given: ΔABC is right triangle in which ∠ABC = 90°.

A circle is drawn on side AB as diameter intersecting AC in P.

PQ is the tangent to the circle when intersects BC in Q.

Construction: Join BP.

Proof:

PQ and BQ are tangents drawn from an external point Q.

⇒ PQ = BQ ....................(i) [Length of tangents drawn from an external point to the circle are equal]

⇒ ∠PBQ = ∠BPQ [In a triangle, equal sides have equal angles opposite to them]

As , it is given that,

AB is the diameter of the circle.

∴ ∠APB = 90° [Angle in a semi-circle is 90°]

∠APB + ∠BPC = 180° [Linear pair]

⇒∠BPC = 180° – ∠APB = 180° – 90° = 90°

In ΔBPC,

∠BPC + ∠PBC + ∠PCB = 180° [Angle sum property]

⇒∠PBC + ∠PCB = 180° – ∠BPC = 180° – 90° = 90° ...................(ii)

Now,

∠BPC = 90°

⇒∠BPQ + ∠CPQ = 90° .................(iii)

From (ii) and (iii), we get,

⇒∠PBC + ∠PCB = ∠BPQ + ∠CPQ

⇒ ∠PCQ = ∠CPQ [∠BPQ = ∠PBQ]

In ΔPQC,

∠PCQ = ∠CPQ

∴ PQ = QC ...(iv)

From (i) and (iv), we get,

BQ = QC

Thus, tangent at P bisects the side BC.

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