Question

I am in 10th class. In ∆ABC, AB=AC. P is the midpoint of AC and Q is the midpoint of AB. Prove that quadrilateral BCPQ is cyclic.​

Answer 1

Step-by-step explanation:

PQ || BC

PQ || BC (The line joining midpoints of two sides of a circle will be parallel to the third side).

PQ || BC (The line joining midpoints of two sides of a circle will be parallel to the third side). ∠B = ∠C......(1)

(1) (∵ AB = AC are given)

(1) (∵ AB = AC are given) ∠C + ∠Q = 180°......(2)

(2) (∵ PQ || BC, QC is the bisector, so the sum of alternate angles are 180°)

(2) (∵ PQ || BC, QC is the bisector, so the sum of alternate angles are 180°) ∠B + ∠Q = 180°.

(2) (∵ PQ || BC, QC is the bisector, so the sum of alternate angles are 180°) ∠B + ∠Q = 180°. From (1) and (2) we conclude that BPQC is an cyclic trapezium.

•Hence proved.