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Question
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
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Solution
Given:
ABCD is a trapezium whose non-parallel sides AD and BC are equal.
Trapezium ABCD is cyclic.
Join BE, where BE || AD
Proof: Since, AB || DE and AD || BE
Since the quadrilateral ABED is a parallelogram.
∴ ∠BAD = ∠BED ...(i) [Opposite angles of a parallelogram are equal]
And AD = BE ...(ii) [Opposite angles of a parallelogram are equal]
But AD = BC [Given] ...(iii)
From equations (ii) and (iii)
BC = BE
⇒ ∠BEC = ∠BCE ...(iv) [Angles opposite to equal sides are equal]
Also, ∠BEC + ∠BED = 180° ...[Linear pair axiom]
∴ ∠BCE + ∠BAD = 180° ...[From equations (i) and (iv)]
If the sum of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.
Hence, trapezium ABCD is cyclic.
Hence proved.
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