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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

ΔPACΔPBC

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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