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Question
In the given figure, AB = AC. Prove that DECB is an isosceles trapezium.
Solution
Here, AB = AC
`=>` ∠B = ∠C
∴ DECB is a cyclic quadrilateral
(In a triangle, angles opposite to equal sides are equal)
Also, ∠B + ∠DEC = 180° ...(1)
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
`=>` ∠C + ∠DEC = 180° ...[From (1)]
But this is the sum of interior angles
On one side of a transversal.
∴ DE || BC
But ∠ADE = ∠AED = ∠C ...[Corresponding angles]
Thus, ∠ADE = ∠AED
`=>` AD = AE
`=>` AB – AD = AC – AE ...(∴ AB = AC)
`=>` BD = CE
Thus, we have, DE || BC and BD = CE
Hence, DECB is an isosceles trapezium
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