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. The radius of the circle is r.

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

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

Similarly,

A(∆APB) = `1/2` × AB × "r"     ......(ii)

A(∆APC) = `1/2` × AC × "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"`