In the given 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

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Solution

In the given problem, the Right Hand Side of all the options is same, that is,

AB + BC + CA

So, we shall find out AB + BC + CA and check which of the options has the Left Hand Side value which we will arrive at.

By looking at the figure, we can write,

AB + BC + CA = AB + BF + FC + CD

We know that tangents drawn from an external point will be equal in length. Therefore,

BF = BE

FC= CD

Now we have,

AB + BC + CA = AB + BE + CD + CA

AB + BC + CD = (AB + BE) + (CD + CA)

By looking at the figure, we write the above equation as,

AB + BC + CD = AE + AD

Since tangents drawn from an external point will be equal,

AE = AD

Therefore,

AB + BC + CD = AD + AD

AB + BC + CD = 2AD

Therefore option (b) is the correct answer.

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In the given 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

In the given 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