Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines
Solution:
Given, the centre of a circle touches two intersecting lines
We have to prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines.
Consider a circle with centre O.
PR and PQ are the tangents drawn through an external point P to the circle.
Join OR and OQ which represents the radius of the circle.
We know that the radius of the circle is perpendicular to the tangent at the point of contact.
So, OR ⟂ PR and OQ ⟂ PQ
Considering triangles PRO and PQO.
PO = PO = common side
OR = OQ = radius
∠PQO = ∠PRO = 90°
By RHS criterion, the triangles PQO and PRO are similar.
By corresponding parts of congruent triangles,
The angles ∠RPO and ∠QPO are equal.
So, ∠RPO = ∠QPO
This implies that O lies on the angle bisector of PR and PQ.
Therefore, it is proved that the centre of the circle lies on the angle bisector of PR and PQ.
✦ Try This: In the adjoining figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the bisector of ∠BAC.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.3 Problem 4
Prove that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines
Summary:
It is proven that the centre of a circle touching two intersecting lines lies on the angle bisector of the lines by corresponding parts of congruent triangles
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