Question

do it... fast..guys​

Answer 1

AF=AE and BD = BF and CE = CD

Since, tangents drawn to the same circle from a common point are congruent

Thus we can easily prove

AF+BD+CE =AE+BF+CD..........(1)

Now consider the third term which we have to prove equal

i.e

$\frac{1}{2} \times (perimeter \: of \: abc)$

Perimeter of ∆ABC

= AB+BC+AC.....general formula

Here AB= AF+FB and similarly

BC= BD+CD

AC= AE+EC

THUS,

PERIMETER =AF+FB+BD+CD+AE+EC

Let me rearrange the terms

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

I proved some values are equal during the start of this answer by using that

PERIMETER= 2AE+2BF+2CD

LET'S divide each term by 2

Perimeter of ∆ABC/2 = AE+BF+CD

Look at the RHS it's same as (1)

Thus we proved

the given three terms equal

Answer 2

Answer:

perimeter of ∆ABC/2 =

= AE + BF + CD

Hope it will be helpful ✌️