Question

In the given figure, the side BC, CA and AB of a triangle ABC touch a circle with centre O and radius r at points D, E and F respectively. Prove that
1) AB+CD=AC+BD
2) Area of triangle ABC = 1/2.r(perimeter of triangle ABC)​

Answer 1

Proved below.

Step-by-step explanation:

Given:

Here the side BC, CA and AB of a triangle ABC touches a circle having center O and radius r at points D, E and F respectively.

We know that, tangents from an external point to the same circle are equal.

Therefore, AF = AE, BF = BD and CD = CE

(1)  Now, adding both the sides, we get

AF + BF + CD = AE + BD + CE

( AF + BF) + CD =( AE + CE) + BD

AB + CD = AC + BD        [∵ AB = AP + PB and AC = AR + RC]

(2) Area (ΔABC) = Area (Δ OBC) + Area (ΔOAC) + Area (ΔOAB)

= $\frac{1}{2}\times BC \times r +\frac{1}{2}\times AC \times r +\frac{1}{2}\times AB \times r$

 =   $\frac{1}{2}[BC + AC + AB ] \times r$

=  $\frac{1}{2} \times (Perimeter \triangle ABC) \times r$

Hence Proved.