Question

A circle with centre $P$ is inscribed in the $△ABC$. Side $AB$, side $BC$ and side $AC$ touch the circle at points $L,M$ and $N$ respectively. Radius of the circle is $r$.
Prove that:
$$\begin{array}{c}A\left(△ABC\right)=\frac{1}{2}(AB+BC\\ +AC)\times r\end{array}$$

Answer 1

Given: Side $AB$, side $BC$ and side $AC$ are tangents to circle at $L,M$ and $N$ respectively. Radius $=1$
To prove: $A\left(△ABC\right)=\frac{1}{2}(AB+BC+AC)\times r$
Construction: Join seg $PM$, seg PN, seg PL, seg AP, seg BP and seg CP.
Proof: seg $BC$ is a tangent to circle at $M$.
$\therefore$ seg $PM\perp segBC\dots \dots$ [ Tangent is perpendicular to radius]
$$\begin{array}{cc}& A\left(△BPC\right)=\frac{1}{2}\times BC\times PM\\ & \therefore A\left(△BPC\right)=\frac{1}{2}\times BC\times r\dots ..\left(i\right)[PM=\text{radius}=r]\end{array}$$
Similarly,
$\begin{array}{cc}& A\left(△APB\right)=\frac{1}{2}\times AB\times r\dots \dots \text{(ii)}\\ & A\left(△APC\right)=\frac{1}{2}\times AC\times r\dots \dots \text{(iii)}\end{array}$
NoW,
$\begin{array}{cc}& A\left(△ABC\right)=A\left(△APB\right)+A\left(△BPC\right)+A\left(△APC\right)\dots \dots [\text{Area addition property]}\\ & =\frac{1}{2}\times AB\times r+\frac{1}{2}\times BC\times r+\frac{1}{2}\times AC\times r\dots \dots [\text{From (i) , (ii) and (iii)]}\\ & =\frac{1}{2}r(AB+BC+AC)\\ & \therefore A\left(△ABC\right)=\frac{1}{2}(AB+BC+AC)\times r\end{array}$