Question

A circle with centre P is inscribed in the ∆ABC. Side AB, side BC and side AC touches the circle at points L, M and N respectively. Radius of the circle is r.

Prove that: `A(ΔABC) = 1/2 (AB + BC + AC) xx r`

Answer 1

Given: Side AB, side BC and side AC are tangents to circle at L, M and N respectively. Radius = r

To prove: `A(∆ABC) = 1/2 (AB + BC + AC) xx 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.

∴ seg PL ⊥ side AB

seg PM ⊥ side BC

seg PN ⊥ side AC   ...(By the theorem, the tangent is perpendicular to the radius)

We know,

Area of triangle = `1/2 xx "base" xx "height"`

∴ seg PM ⊥ seg BC   ...[Tangent is perpendicular to radius]

`A(∆BPC) = 1/2 xx BC xx PM`

∴ `A(∆BPC) = 1/2 xx BC xx r`   ...(i) [PM = radius = r]

Similarly,

`A(∆APB) = 1/2 xx AB xx r`   ...(ii)

`A(∆APC) = 1/2 xx AC xx r`   ...(iii)

Now,

A(∆ABC) = A(∆APB) + A(∆BPC) + A(∆APC)   ...[Area addition property]

= `1/2 xx AB xx r + 1/2 xx BC xx r + 1/2 xx AC xx r`   ...[From (i), (ii) and (iii)]

= `1/2 xx r (AB + BC + AC)`

∴ `A(∆ABC) = 1/2 (AB + BC + AC) xx r`