Question

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

Hint:

The segment \(PM\) is called the semi-perimeter of triangle \(PQR\). In such configurations involving excircles, the distance from the vertex to the point of contact on the extended side always equals the semi-perimeter.

Answer 1

Step 1: Understanding the Concept:
The lengths of tangents drawn from an external point to a circle are equal. We will identify these points and their corresponding tangents.
Step 2: Key Formula or Approach:
From point \(P\): \(PM = PN\)
From point \(Q\): \(QM = QS\)
From point \(R\): \(RN = RS\)
Step 3: Detailed Explanation:
Let's find the perimeter of \(\Delta PQR\):
\[ \text{Perimeter} = PQ + QR + PR \]
We can split side \(QR\) at point \(S\):
\[ = PQ + (QS + SR) + PR \]
Substituting equal tangents (\(QS = QM\) and \(SR = RN\)):
\[ = PQ + QM + RN + PR \]
Observing the segments: \(PQ + QM = PM\) and \(PR + RN = PN\).
\[ = PM + PN \]
Since tangents from external point \(P\) are equal (\(PM = PN\)):
\[ \text{Perimeter} = PM + PM = 2PM \]
Therefore:
\[ PM = \frac{1}{2}(\text{Perimeter of } \Delta PQR) = \frac{1}{2}(PQ + QR + PR) \]
Step 4: Final Answer:
Hence proved.