In the isosceles triangle ABC, AB = AC. A straight line parallel to BC intersects AB and AC at points P and Q respectively. Prove that the quadrilateral BCQP is cyclic (i.e., its vertices lie on a circle).

Q.In the isosceles triangle ABC, AB = AC. A straight line parallel to BC intersects AB and AC at points P and Q respectively. Prove that the quadrilateral BCQP is cyclic (i.e., its vertices lie on a circle).

Let ABC be an isosceles triangle where AB = AC. A line parallel to BC intersects AB and AC at points P and Q respectively. We need to prove that quadrilateral BCQP is cyclic. Proof: Since AB = AC, \[ \Rightarrow \angle ABC = \angle ACB \Rightarrow \angle PBC = \angle QCB \] Also, since PQ is parallel to BC, \[ \Rightarrow \angle APQ = \angle PBC \text{and} \quad \angle AQP = \angle QCB \Rightarrow \angle PBC = \angle QCB = \angle APQ = \angle AQP \] Now, in quadrilateral BCQP: \[ \angle BPQ + \angle BCQ = \angle BPQ + \angle APQ = 180^\circ \text{and} \quad \angle PQC + \angle PBC = \angle PQC + \angle AQP = 180^\circ \] Therefore, the opposite angles of quadrilateral BCQP are supplementary. BCQP is a cyclic quadrilateral.