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If the non-parallel sides of a trapezium...

If the non-parallel sides of a trapezium are equal, then

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Given ABCD is a trapezium whose non-parallel sides AD and BC are equal.

To prove Trapezium ABCD is a cyclic.
Join BE, where `BE||AD`.
Proof Since, `AB|| DE and AD||BE`
Since, the quadrilateral ABCD is a parallelogram.
`:. angleBAD=angleBED` ...(i)
[opposite angles of a parallelogram are equal]
and AD=BC ..(ii)
[opposite sides of a parallelogram are equal]
But AD=BC [given]...(iii)
From Eqs. (ii) and (iii), BC=BE
`rArr angleBEC=angleBCE` .....(iv)
[angles opposite to equal sides are equal]
Also, `angleBEC=angleBED=180^(@)` [linear pair axiom]
`:. angleBCE+angleBAD=180^(@)` [from Eqs. (i) and (iv)]
If sum of opposite angles of a quaderilateral is `180^(@)`, then quadrilateral is cyclic.
Hence, trapezium ABCD is a cyclic.
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