Question

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

• A. $AD=AB+BC+CA$
• B. $2AD=AB+BC+CA$
• C. $3AD=AB+BC+CA$
• D. $4AD=AB+BC+CA$

Answer 1

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),

$AD+AD=AB+BF+CF+AC$

$2AD=AB+BC+AC$