ABC is an isosceles triangle where AB = AC which is circumscribed about a circle. If P is the point where the circle touches the side BC, then which of the following is true?

The correct option is BP = PC.

Let us understand the geometric properties of a circle inscribed in a triangle step-by-step.

Step 1: Identify the points of contact
Let the circle touch the sides $AB$, $BC$, and $AC$ of triangle $ABC$ at the points $Q$, $P$, and $R$ respectively.

Step 2: Apply the Tangent Theorem
According to the properties of circles, the lengths of two tangents drawn from an external point to a circle are equal. Applying this theorem to the vertices of triangle $ABC$:
1. From vertex $A$, the tangents are $AQ$ and $AR$. Therefore:

$AQ=AR$

2. From vertex $B$, the tangents are $BQ$ and $BP$. Therefore:

$BQ=BP$

3. From vertex $C$, the tangents are $CR$ and $CP$. Therefore:

$CR=CP$

Step 3: Utilize the Isosceles Triangle Property
We are given that $ABC$ is an isosceles triangle where:

$AB=AC$

The side $AB$ can be written as the sum of its segments $AQ+BQ$, and the side $AC$ can be written as the sum of its segments $AR+CR$. Thus:

$AQ+BQ=AR+CR$

Step 4: Solve for the relationship between BP and PC
Since we know from Step 2 that $AQ=AR$, we can subtract this equal length from both sides of the equation:

$BQ=CR$

Now, substituting the tangent relationships $BQ=BP$ and $CR=CP$ into this equation gives:

$BP=CP$

Or equivalently:

$BP=PC$

Thus, the point of contact $P$ bisects the side $BC$, proving that $BP=PC$ is true.