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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.
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Solution
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°.
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