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.

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.

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