A circle is touching the side BC of a `triangle`ABC at point P and touching AB and AC produced at Q and R respectively. Prove that `AQ=(1)/(2)("perimeter of "triangle ABC).`

Explanation

To prove that \( AQ = \frac{1}{2} \) (perimeter of triangle \( ABC \)), we will follow these steps:

### Step 1: Define the points and segments
Let \( A, B, C \) be the vertices of triangle \( ABC \). The incircle of the triangle touches side \( BC \) at point \( P \), and it touches the extensions of sides \( AB \) and \( AC \) at points \( Q \) and \( R \) respectively.

### Step 2: Use the properties of tangents
From the properties of tangents to a circle, we know that the lengths of tangents drawn from an external point to a circle are equal. Therefore, we can write:
- \( AQ = AP \) (let this be equation 1)
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