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Question

In the adjoining figure, if AD, AE and BC are tangents to the circle at D, E and F respectively. Then


A

AD=AB+BC+CA

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B

2AD=AB+BC+CA

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C

3AD=AB+BC+CA

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D

4AD=AB+BC+CA

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Solution

The correct option is B

2AD=AB+BC+CA



We know that

AD=AE

AD=AB+BE

Since BE=BF as tangents drawn from an external point to a circle are equal , AD=AB+BF……(1)

Also

AD=AC+CD

AD=AC+CF ……(2) (CD and CF are tangents drawn from the external point C to the circle)

Adding equation (1) and (2),

AD+AD=AB+BF+CF+AC

2AD=AB+BC+AC


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