Proof:

Let's assume that the triangle ABC is isosceles, with AB = AC. We need to prove that BP > PQ, where P lies on the angle bisector of BAC and Q is the intersection point of the extended line BP and AC.

1. Construction:
Construct the angle bisector of BAC and label it as AD, where D is the point of intersection with BC.
Draw a line parallel to AD from point B, intersecting AC at point Q.

2. Proof:
Since AD is the angle bisector of BAC, we can conclude that angles BAD and CAD are equal.
Therefore, triangles ABD and ACD are congruent by the SAS (side-angle-side) congruence criterion.
Hence, AD = AD (common side), AB = AC (given), and angles BAD and CAD are equal.

3. Property 1:
We can observe that triangles ABP and ACQ are similar by the AA (angle-angle) similarity criterion.
Both triangles have a common angle at A, and the angles at B and C are equal because they are corresponding angles when AB is parallel to CQ.

4. Property 2:
From the similarity of triangles ABP and ACQ, we can write the following proportion:
AB / AC = BP / CQ

Since AB = AC (given), the above proportion simplifies to:
1 = BP / CQ

5. Property 3:
Considering the fact that CQ is the extended line segment of BP, we can conclude that CQ > BP.
Hence, 1 = BP / CQ < 1,="" which="" implies="" that="" bp="" />< />

6. Conclusion:
From Property 3, we can say that BP < cq,="" which="" means="" that="" bp="" is="" always="" less="" than="" />
Therefore, we have proved that BP > PQ, as PQ is a part of CQ.

Hence, the statement is true for any point P lying on the angle bisector of BAC in an isosceles triangle ABC with AB = AC.

Hby