Question

The in circle of a triangle ABC touches the sides AB,BC,CA at the points F,D and E respectively ,show that AF+BD+CE=FB+DC+EA=1/2 perimeter of triangle ABC

Answer 1

If you would have provided the figure, it would have been easy to answer...

$\huge\tt\orange{♬•Answer:}$

Let ABC be a triangle and the points at which circle is touching the triangle be DEF.

Now, We know that the lengths of tangents from an exterior point to a circle are equal

∴

• AF=AE (i)
• BD=BF (ii)
• CE=CD (iii)

Therefore,adding i,ii and iii,

AF+BD+CE=AE+BF+CD (let it be equal to k)

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

Now,Perimeter of ΔABC

=AB+BC+CA

Now,AB can be written as AF+BF

BC can be written as BD+CD

CA can be written as CE+AE

Now,

AB+BC+CA=AF+BD+CE+AE+BF+CD

AB+BC+CA=k+k (see iv)

AB+BC+CA=2k

Perimeter=2k

k=Perimeter/2

AF+BD+CE=EA+FB+DC =Perimeter/2 ( see iv again)