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

$A D = A B + B C + C A$

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$2 A D = A B + B C + C A$

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$3 A D = A B + B C + C A$

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$4 A D = A B + B C + C A$

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Solution

The correct option is B
$2 A D = A B + B C + C A$

We know that

$A D = A E$ [Tangents from an external point to the circle are equal in length]

$\Rightarrow A D = A B + B E$

$A D = A B + B F$ [since, $B E = B F$ ] ……(1)

Also,

$A D = A C + C D$

$\Rightarrow A D = A C + C F$ [since, $C D = C F$ ] ……(2)

Adding equations (1) and (2)

$A D + A D = A B + B F + C F + A C$

$\Rightarrow 2 A D = A B + A C + B C$

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