Question

In the given figure, if a circle touches the side QR of ΔPQR at S and extended sides PQ and PR at M and N, respectively, then prove that PM = `1/2` (PQ + QR + PR)

Answer 1

Given: A circle is touching a side QR of ΔPQR at point S.

PQ and PR are produced at M and N respectively.

To prove: PM = `1/2` (PQ + QR + PR)

Proof: PM = PN  ...(i) (Tangents drawn from an external point P to a circle are equal)

QM = QS  ...(ii) (Tangents drawn from an external point Q to a circle are equal)

RS = RN  ...(iii) (Tangents drawn from an external point R to a circle are equal)

Now, 2PM = PM + PM

= PM + PN  ...[From equation (i)]

= (PQ + QM) + (PR + RN)

= PQ + QS + PR + RS  ...[From equations (i) and (ii)]

= PQ + (QS + SR) + PR

= PQ + QR + PR

∴ PM = `1/2` (PQ + QR + PR)

Hence proved.