Question

Let ‘s’ denote the semiperimeter of a triangle ABC in which $$BC=a$$ , $$CA=b$$ , $$AB=c$$ . If the circle touches the sides BC, CA, AB at D, E, F respectively, prove that $$BD=s-b$$ .

Answer 1

Hint: We know that the semi perimeter of a triangle is given by $$s={\displaystyle \frac{a+b+c}{2}}$$ , where ‘a’ ‘b’ and ‘c’ are the length of sides of the triangle. We know the theorem that the length of tangents drawn from an external point to a circle are equal. Using this theorem we can solve the above theorem and we will draw a diagram using the given data in the above problem.

Complete step by step solution:
Let’s draw the diagram for the given data.
 $$BC=a$$ , $$CA=b$$ , $$AB=c$$ .
Also the circle touches the sides BC, CA, AB at D, E, F respectively.

So the semi-perimeter is given according to the question is
 $$s={\displaystyle \frac{a+b+c}{2}}$$
 $$\Rightarrow 2s=a+b+c$$ .
B is an external point and BD and BF are tangents and from an external point the tangents drawn to a circle are equal in length.
That is,
 $$BD=BF---(1)$$ .
(The length of tangents drawn from an external point to a circle are equal.)
Similarly we have,
 $$AF=AE---(2)$$ and $$CD=CE---(3)$$ .
Also given ‘s’ is semi perimeter
 $$s={\displaystyle \frac{AB+AC+BC}{2}}$$
 $$\Rightarrow 2s=AB+AC+BC---(4)$$
From the diagram we have,
 $$AB=AF+FB$$
 $$AC=AE+EC$$
 $$BC=BD+DC$$
Substituting these in equation (1) we have,
 $$\Rightarrow 2s=AF+FB+AE+EC+BD+DC$$
Now using equation (1), (2) and (3) we have,
 $$\Rightarrow 2s=2AE+2EC+2BD$$
 $$\Rightarrow 2s=2(AE+EC+BD)$$
 $$\Rightarrow s=AE+EC+BD$$
But $$AE+EC=AC$$ . (See in the above diagram)
 $$\Rightarrow s=AC+BD$$
But we know that $$AC=b$$
 $$\Rightarrow s=b+BD$$
 $$\Rightarrow s-b=BD$$
Rearranging we have,
 $$\Rightarrow BD=s-b$$ .
Hence proved.

Note: We know that if we want perimeter we add all the side lengths of a given dimension. We know that the perimeter of a triangle is $$P=a+b+c$$ , where ‘a’, ‘b’ and ‘c’ are the length of sides of the triangle. Since they asked for a semi we took half of the perimeter. To solve we need to remember particular theorems.