Question

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.

Answer 1

Given: ΔABC is an isosceles triangle such that AB = AC and also DE || BC.

To prove: Quadrilateral BCDE is a cyclic quadrilateral.

Construction: Draw a circle passes through the points B, C, D and E.

Proof: In ΔABC, AB = AC   ...[Equal sides of an isosceles triangle]

⇒ ∠ACB = ∠ABC   ...(i)

Since, DE || BC  ...[Angles opposite to the equal sides are equal]

⇒ ∠ADE = ∠ACB  [Corresponding angles] ...(ii)

On adding both sides by ∠EDC in equation (ii), we get

∠ADE + ∠EDC = ∠ACB + ∠EDC

⇒ 180° = ∠ACB + ∠EDC  ...[∠ADE and ∠EDC from linear pair axiom]

⇒ ∠EDC + ∠ABC = 180°  ...[From equation (i)]

Hence, BCDE is a cyclic quadrilateral, because sum of the opposite angles is 180°.