Prove that the angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.
Solution:
Consider a triangle ABC
We have to prove that the angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ΔABC.
Let the angle bisector of ∠A intersect circumcircle of ΔABC at P.
Join BP and CP.
We know that the angles in the same segment of a circle are equal
So, ∠BAP = ∠BCP
Since, AP is bisector of ∠A.
∠BAP = ∠BCP
∠A = ∠BAP + ∠BCP
So, ∠BAP = ∠BCP = 1/2 ∠A -------------------------- (1)
Similarly, ∠PAC = ∠PBC = 1/2 ∠A ------------------- (2)
From (1) and (2),
∠BCP = ∠PBC
We know that, if the angles subtended by two Chords of a circle at the centre are equal, then the chords are equal.
So, BP = CP
P lies on the perpendicular bisector of BC.
Therefore, angle bisector of ∠A and perpendicular bisector of BC intersect on the circumcircle of ΔABC.
✦ Try This: ABC is a triangle with ∠A > ∠C and D is a point on BC such that ∠BAD = ∠ACB. The perpendicular bisectors of AD and DC intersect in the point E. Prove that ∠BAE = 90°.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.4 Problem 5
Prove that the angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.
Summary:
It is proven that an angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if they intersect, they will intersect on the circumcircle of the triangle
☛ Related Questions:
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- In Fig. 10.19, AB and CD are two chords of a circle intersecting each other at point E. Prove that ∠ . . . .