Question

Sides AB BC AND CA OF TRIANGLE ABC TOUCH THE CIRCLE WITH CENTRE O AND RADIUS r AT P Q AND R RESPECTIVELY PROVE THAT AREA OF TRIANGLE ABC IS HALF THE PERIMETER OF TRIANGLE ABC INTOr

Answer 1

Given: ΔABC with circle inscribed within it. P, Q, R the points of contact for tangents AB, BC, & CA. Inradius is r. Semiperimeter is s.

RTP: ar(ΔABC) = sr

Proof: AP=AR, BP=BQ, & CQ=CR due to tangent theorem. Let AP=AR=x, BP=BQ=y, & CQ=CR=z.

Angles at P, Q, & R, are 90° according to tangent theorem.

So, ar(ΔABC) = ar(ΔAPO)+ar(ΔARO)+ar(ΔBPO)+ar(ΔBQO)+ar(ΔCRO)+ar(ΔCQO)

So ar(ΔABC) = 1/2(rx) + 1/2(rx) + 1/2(rz) + 1/2(rz) + 1/2(ry) + 1/2(ry)

So ar(ΔABC) = rx + rz + ry = r(x+y+z)

Semiperimeter = 1/2(AB + BC + CA) = 1/2(x + z + z + y + x + y) = 1/2(2x + 2y + 2z) = x + y + z

Thus, ar(Δ) = rs