Question

In the following figure, the inscribed circle of $\Delta ABC$ with centre P touches the sides AB, BC and AC at points L, M, N respectively. Show that $A\left(\Delta ABC\right)=\frac{1}{2}\times$ (perimeter of $\Delta ABC$) $\times$ (radius of inscribed circle).

Answer 1

Given: The inscribed circle of $\Delta ABC$ touches side AB at L, side BC at M and side AC at N.

Proof: By Tangent $\perp$ theorem
Radius PL $\perp$ AB, PM $\perp$ BC and PN $\perp$ AC
In $\Delta APBA\left(\Delta APN\right)=\frac{1}{2}\times b\times h$
$=\frac{1}{2}\times AB\times r$ ....... (1)
In $\Delta BPCA\left(\Delta BPC\right)=\frac{1}{2}\times BC\times r$ ....... (2)
In $\Delta APCA\left(\Delta APC\right)=\frac{1}{2}\times AC\times r$ ........ (3)
Adding (1), (2) and (3)
$A\left(\Delta APB\right)+A\left(\Delta BPC\right)+A\left(\Delta APC\right)=\frac{1}{2}\times AB\times r+\frac{1}{2}\times BC\times r+\frac{1}{2}\times AC\times r$
$=\frac{1}{2}(AB+BC+AC)\times r$
$\therefore A\left(\Delta ABC\right)=\frac{1}{2}\left(\text{perimeter of}\Delta ABC\right)\times \text{radius of inscribed circle}$