Question

Question 2
If non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer 1

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

To prove that trapezium ABCD is cyclic.
Join BE, such that BE || AD.

Proof
Since AB || DE and AD || BE, the quadrilateral ABED is a parallelogram.
$\angle BAD=\angle BED$ [opposite angles of a parallelogram are equal ]...............(i)
AD = BE [opposite sides of a parallelogram are equal ] ...............(ii)

Also, AD = BC [Given].............................(iii)
From Eqs. (ii) and (iii),
BC = BE
$\Rightarrow \angle BEC=\angle BCE$ [angles opposite to equal sides are equal]...............(iv)
$Also,\angle BEC+\angle BED={180}^{\circ }$ [ linear pair]
$\therefore \angle BCE+\angle BAD={180}^{\circ }$ [From (i) and (iv)]
If sum of opposite angles of a quadrilateral is ${180}^{\circ }$, then quadrilateral is cyclic.
Therefore, trapezium ABCD is cyclic.