Question

In the given figure, if a circle touches the side QR of Δ PQR 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: PM, PN, QR are tangents.

Length of tangent from external point to a circle is equal:

PM = PN    ...(i)

QM = QS    ...(ii)

RN = RS    ...(iii)

The perimeter of the triangle is the sum of its three sides:

Perimeter = PQ + QR + PR

Since the point S lies on QR,

QR = QS + SR    ...(iv)

The point S  lies on Q, PM, 

PM = PQ + QM   ...(v)

The point S  lies on R, PN, 

PN = PR + RN   ...(vi)

RHS = `1/2` [PQ + QR + PR]

= `1/2`[PQ + QS + SR + PR]    ...[Using equation (iv)]

= `1/2`[PQ + QM + RN + PR]    ...[Using equations (ii) and (iii)]

= `1/2`[PM + PN]    ...[Using equations (v) and (vi)]

= `1/2` [PM + PM]    ...[Using equation (i)]

= `1/2 xx 2 PM`

= PM

= LHS.

Hence proved.