Question

Prove that the tangents at the extremities of any chord make equal angles with the chord. [3 MARKS]

Solution

Concept: 1 Mark

Application: 2 Marks

Let AB be a chord of a circle with center O, and let AP and BP be the tangents at A and B respectively.

Suppose the tangents meet at P .

Join OP. Suppose OP meets AB at C . We have to prove that ∠PAC=∠PBC .

In triangles PCA and PCB , We have

PA = PB

∠APC=∠BPC

[The tangents are equally inclined to line joining extrernal point and centre of circle.]

PC = PC [Common]

So, by SAS - criterion of congruence, we have

ΔPAC≅ΔPBC

⇒∠PAC=∠PBC

Alternative Solution:

In ΔPAB, PA=PB ( ∵ tangents drawn from an external point to a circle are equal.)

∵ PA = PB, ∠PAB = ∠ PBA (Equal opposite sides make equal angles)

⇒ ∠PAC = ∠PBC

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