Question

A circle touches side $BC$ at point $P$ of $\triangle ABC$, from outside of the triangle. Further extended lines $AC$ and $AB$ are tangents to the circle at $N$ and $M$ respectively. Prove that: 
\[ AM = \frac{1}{2} (\text{Perimeter of } \triangle ABC) \] 

Hint:

The tangents drawn from an external point to a circle are equal in length. Use this property to relate the perimeter of the triangle with the tangent segment lengths.

Answer 1

Step 1: Given. 
A circle touches the sides of $\triangle ABC$ externally at points $P$, $M$, and $N$ such that the tangents from a single external point to a circle are equal in length. 
Step 2: Tangent length properties. 
Let the tangents drawn from each vertex be as follows: 

Step 3: Express the sides of the triangle. 

Step 4: Find the perimeter of the triangle. 
\[ \text{Perimeter of } \triangle ABC = AB + BC + CA = (x + y) + (y + z) + (z + x) \] \[ \text{Perimeter} = 2(x + y + z) \] Step 5: Relation of $AM$. 
From the figure, $AM = x$. 
Hence, \[ x + y + z = \frac{1}{2} (\text{Perimeter of } \triangle ABC) \] Therefore, \[ AM = \frac{1}{2} (\text{Perimeter of } \triangle ABC) \] Hence proved. 
Result: $AM = \dfrac{1}{2} (\text{Perimeter of } \triangle ABC)$