Question

In the 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

Answer: (B)

$2AD=AB+BC+CA$

Given-

$AD,AE$ are tangents to a circle from $A$ at D & E, respectively.

Also, $BC$ is another tangent which touches the circle at $F$ and meets $AD\&AE$ at $C\&B$ respectively.

To find out- which of the options is true.

Solution-

We know that the lengths of the tangents, from a point to a circle, are equal.

$\therefore AD=AE,CD=CF\&BE=BF.......\left(i\right)$
Now $AD=AC+CD=AC+CF\&AE=AD=AB+BE=AB+BF......\left(ii\right)$.
Adding $\left(i\right)\&\left(ii\right)$ we get
$2AD=AC+CF+AB+BF=AC+AB+(BF+CF)=AC+AB+BC=AB+BC+CA.$
So $2AD=AB+BC+CA.$
Ans- option B.