"Question 10 In any triangle ABC, if the angle bisector of ∠A and perpendicular bisector of BC intersect, prove that they intersect on the circum circle of the triangle ABC.
Class 9 - Math - Circles Page 187"
Answers
The region between a chord and either of its arcs is called a segment the circle.
Angles in the same segment of a circle are equal.
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Let bisectors of ∠A meet the circumcircle of ∆ABC at M.
Join BM & CM
∠MBC = ∠MAC
[Angle in the same segment are equal]
∠BCM = ∠BAM
[Angle in the same segment are equal]
But , ∠BAM= ∠CAM..........(i)
[Since, AM is bisector of ∠A]
∠MBC= ∠BCM
MB= MC
[SIDE OPPOSITE TO EQUAL ANGLES ARE EQUAL]
So,M must lie on the perpendicular bisector of BC.
Let, M be a point on the perpendicular bisector of BC which lies on circumcircle of ∆ABC. Join AM.
Since, M lies on perpendicular bisector of BC.
BM= CM
∠MBC= ∠MCB
But ∠MBC= ∠MAC
[Angles in same segment are equal]
∠MCB= ∠BAM
[Angles in same segment are equal]
From eq i
∠BAM= ∠CAM
So, AM is the bisector of ∠A
Hence, bisector of ∠A and perpendicular bisector BC intersect at M which lies on the circumcircle of the ∆ABC..
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Hope this will help you....
Answer of this question is following
