Q. 37

# In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Answer :

Given: Bisector AP of angle A of ΔABC and the perpendicular bisector PQ of its opposite side BC intersect at P.

To Prove: P lies on the circumcircle of the triangle ABC.

Construction: Draw the circle through three non-collinear points A, B and P.

Proof: ∠ BAP = ∠ CAP

⇒ chord BP = chord CP

⇒ BP = CP

In Δ BMP and Δ CMP,

BM = CM | M is the bisector of BC

BP = CP |Proved above

MP = MP | Common side

∴ Δ BMP ≅ CMP | SSS

∴ ∠ BMP = ∠CMP | CPCT

But ∠ BMP + ∠ CMP = 180° | Linear Pair Axiom

∴ ∠ BMP = ∠ CMP = 90°

⇒ PM is the right bisector of BC.

Alieter:

Assume that C does not lie on the circle though A, B and P. Let this circle intersect the side AC at C'. (say)

∠ APB = ∠ ACB |Given

∠ APB = ∠ AC'B | Angles in the same segment

∴ ∠ ACB = ∠ AC'B

⇒ C and C' coincide

⇒ The assumption that the point C does not lie on the circle is false.

∴ A, B, P and C are concyclic.

To Prove: P lies on the circumcircle of the triangle ABC.

Construction: Draw the circle through three non-collinear points A, B and P.

Proof: ∠ BAP = ∠ CAP

⇒ chord BP = chord CP

⇒ BP = CP

In Δ BMP and Δ CMP,

BM = CM | M is the bisector of BC

BP = CP |Proved above

MP = MP | Common side

∴ Δ BMP ≅ CMP | SSS

∴ ∠ BMP = ∠CMP | CPCT

But ∠ BMP + ∠ CMP = 180° | Linear Pair Axiom

∴ ∠ BMP = ∠ CMP = 90°

⇒ PM is the right bisector of BC.

Alieter:

Assume that C does not lie on the circle though A, B and P. Let this circle intersect the side AC at C'. (say)

∠ APB = ∠ ACB |Given

∠ APB = ∠ AC'B | Angles in the same segment

∴ ∠ ACB = ∠ AC'B

⇒ C and C' coincide

⇒ The assumption that the point C does not lie on the circle is false.

∴ A, B, P and C are concyclic.

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