Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord
Solution:
We have to prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.
From the figure,
O is the centre of the circle.
AB is a chord of the circle.
PA and PB are the tangents drawn at the ends of a chord of the circle.
The tangents intersect at P.
Join OA and OB.
OA = OB = radius of the circle.
Considering triangle OAB,
OA = OB = radius of the circle.
Since the two sides of the triangle are equal, OAB is an isosceles triangle.
We know that the angles opposite to the equal sides of a triangle are equal.
So, ∠OAB = ∠OBA
i.e., ∠1 = ∠2 ------------- (1)
We know that the radius of the circle is perpendicular to the tangent at the point of contact.
So, OA ⟂ PA and OB ⟂ PB
∠OAP = ∠OBP = 90°
i.e., ∠2 + ∠3 = ∠1 + ∠4 -------------- (2)
Substituting (1) in (2),
∠1 + ∠3 = ∠1 + ∠4
∠1 + ∠3 - ∠1 = ∠4
∠3 = ∠4
Therefore, ∠PAB = ∠PBA.
✦ Try This: It two equal chords of a circle intersect within the circle. Prove that the line joining the point of intersection to the centre makes equal angles with the chords.
☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 10
NCERT Exemplar Class 10 Maths Exercise 9.3 Problem 9
Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord
Summary:
It is proven that the tangents drawn at the ends of a chord of a circle make equal angles with the chord
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