Question

The inscribed circle of ∆ABC with centre P touches the sides AB,BC,AC at points L,M,N respectively
show that A(∆ABC)= 1 upon 2 ( perimeter of ∆ABC) ×(radius of inscribed circle)​

Answer 1

Given: In circle of the ΔABC touches the sides BC, CA and AB at D, E and F respectively.

To prove: AF + BD + CE = AE + BF + CD = (Perimeter of ΔABC)

Proof:

We know that, the length of tangents drawn from and external point to a circle are equal.

∴ AF = AE ...(1)

BD = BF ...(2)

CE = CD ...(3)

Adding (1), (2) and (3), we get

AF + BD + CE = AE + BF + CD ...(4)

Perimeter of ΔABC

= AB + BC + CA

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

= (AF + BD) + (BD + CE) + (CE + AF) [Using (1), (2) and (3)]

= 2(AF + BD + CE)

∴ AF + BD + CE = (Perimeter of ΔABC) ...(5)

From (4) and (5), we get

AF + BD + CE = AE + BF + CD = (Perimeter of ΔABC)