AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.
Solution:
Given, AB and AC are two equal chords of a circle
We have to prove that the bisector of the angle BAC passes through the centre of the circle.
Considering triangles APB and APC,
Given, AB= AC
∠BAP = ∠CAP
Common side = AP
SAS criterion states that if two sides of one triangle are proportional to two sides of another triangle and their included angles are congruent, then the triangles are similar.
By SAS criterion, the triangles APB and APC are similar.
The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are similar, then their corresponding sides and angles are also congruent or equal in measurements.
By CPCTC,
BP = CP
∠APB = ∠APC ---------- (1)
We know that the linear pair of angles is always equal to 180 degrees.
∠APB + ∠APC = 180°
From (1),
∠APB + ∠APB = 180°
2∠APB = 180°
∠APB = 180°/2
∠APB = 90°
This implies AP is the perpendicular bisector of chord BC.
Therefore, AP passes through the center of the circle.
✦ Try This: If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 4
AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.
Summary:
AB and AC are two equal chords of a circle. It is proven that the bisector of the angle BAC passes through the centre of the circle
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