(B) In the given figure, if a circle touches the side QR of $\triangle PQR$ at S and extended sides PQ and PR at M and N respectively, then prove that : $PM = \frac{1}{2}(PQ + QR + PR)$ .

Q: 29. (B) In the given figure, if a circle touches the side QR of $\triangle PQR$ at S and extended sides PQ and PR at M and N respectively, then prove that : $PM = \frac{1}{2}(PQ + QR + PR)$.

Proved: $PM = \frac{1}{2}(PQ + QR + PR)$. Using equal tangents from each external point, $PM = PN$, $QM = QS$ and $RN = RS$. Then $PM = PQ + QM = PQ + QS$, and $PN = PR + RN = PR + RS$. Adding these and using $QS + RS = QR$ gives $PM + PN = PQ + QR + PR$. Since $PM = PN$, this becomes $2\cdot PM = PQ + QR + PR$, so PM equals half the perimeter of $\triangle PQR$.

Circles

Tangents from an external point

Official Answer

Proved: $PM = \frac{1}{2}(PQ + QR + PR)$ . Using equal tangents from each external point, $PM = PN$ , $QM = QS$ and $RN = RS$ . Then $PM = PQ + QM = PQ + QS$ , and $PN = PR + RN = PR + RS$ . Adding these and using $QS + RS = QR$ gives $PM + PN = PQ + QR + PR$ . Since $PM = PN$ , this becomes $2\cdot PM = PQ + QR + PR$ , so PM equals half the perimeter of $\triangle PQR$ .

tangents from external pointequal tangentsexcirclePM = PNQM = QSRN = RSsemiperimeterperimeter of triangle

Marking Scheme

11 mark: Correctly stating the three equal-tangent relations $PM = PN$ , $QM = QS$ and $RN = RS$ with the reason 'tangents from an external point are equal'.
21 mark: Writing $PM = PQ + QS$ and $PN = PR + RS$ and adding to get $PM + PN = PQ + QR + PR$ (using $QS + RS = QR$ ).
31 mark: Concluding $2\cdot PM = PQ + QR + PR$ , hence $PM = \frac{1}{2}(PQ + QR + PR)$ . Full marks for any equivalent correct chain of reasoning.

Hint

Apply 'tangents from an external point are equal' at P, Q and R ( $PM=PN, QM=QS, RN=RS$ ), add PM and PN, and use $QS+RS=QR$ .

Quick Oral Answer

Tangents from a point are equal, so $PM=PN, QM=QS, RN=RS$ . Then $PM=PQ+QS$ and $PN=PR+RS$ ; adding and using $QS+RS=QR$ gives $2PM=PQ+QR+PR$ , so PM is half the perimeter.

Analysis & Explanation

A higher-order proof that applies the 'tangents from an external point are equal' theorem three times, once from each vertex of the triangle.

Concept

The circle is an excircle of $\triangle PQR$ : it touches side QR directly at S, and touches the other two sides only after they are extended, at M and N.
Key tangent equalities: $PM = PN, QM = QS, RN = RS$ .
The decisive step is recognising $QS + RS = QR$ , since S lies between Q and R on side QR.

Common mistakes

Failing to recognise the excircle configuration, since the tangent points M and N lie on the extensions of PQ and PR, not on the sides themselves.
Missing that QS + RS collapses to QR, which is needed to reduce the sum to the perimeter.

Real-world relevance

Excircle tangent lengths underpin Heron's formula ( $\text{Area} = r \cdot s$ ) and are used in triangle geometry, GPS trilateration, and designing circular fillets tangent to extended edges in CAD.

Common Mistakes

1Not recognising that M and N lie on the produced (extended) sides, so writing $PM = PQ - QM$ instead of $PM = PQ + QM$ .
2Forgetting the key substitution $QS + RS = QR$ , which is what turns the sum into the perimeter.
3Stating the equal-tangent relations without giving the reason ('tangents from an external point are equal'), which loses the justification mark.

Interesting Facts

The circle in this problem is an 'excircle' — a triangle has three excircles, each lying outside the triangle opposite one vertex, in addition to its single incircle.

The tangent length from a vertex to the excircle opposite it always equals the semiperimeter $s = (a+b+c)/2$ — exactly the result this question proves.

The equal-tangent theorem is the same principle that lets a belt or chain wrap two pulleys with equal straight-segment lengths on each side — a fact used in mechanical drive design.

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Frequently Asked Questions

What kind of circle is described in this problem?

It is an excircle (escribed circle) of triangle PQR. It touches one side (QR) of the triangle directly and touches the other two sides only when they are extended beyond Q and R. Every triangle has three such excircles.

Why does PM equal half the perimeter?

Because $PM = PN$ (tangents from P), and adding $PM + PN$ reproduces every side of the triangle exactly once ( $PQ + QR + PR$ ). So $2\cdot PM$ equals the full perimeter, making PM the semiperimeter, i.e. half the perimeter.

Which single theorem is used throughout the proof?

The theorem that the lengths of tangents drawn from an external point to a circle are equal. It is applied three times — from P ( $PM = PN$ ), from Q ( $QM = QS$ ) and from R ( $RN = RS$ ).