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

$AD=AE$ [Tangents from an external point to the circle are equal in length]

$\Rightarrow AD=AB+BE$

$AD=AB+BF$ [since, $BE=BF$] ……(1)

Also,

$AD=AC+CD$

$\Rightarrow AD=AC+CF$ [since, $CD=CF$] ……(2)

Adding equations (1) and (2)

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

$\Rightarrow 2AD=AB+AC+BC$