If two non-parallel sides of a trapezium are equal, it is cyclic. OR An isosceles trapezium is always cyclic.

Explanation

Given: `ABCD` is a trapezium where non-parallel sides `AD` and `BC` are equal.
Construction: `DM` and `CN` are perpendicular drawn on `AB` from `D` and `C`, respectively.
To prove: `ABCD` is cyclic trapezium.
Proof: In `/_\DAM` and `/_\CBN`,
`AD=BC` [Given]
`/_AMD=/_BNC` [Right angles]
`DM=CN` [Distance between the parallel lines]
`therefore /_\DAM~=/_\CBN` [By `RHS` congruence condition]
Now, `/_A=/_B` [By `CPCT`]
Also, `/_B+/_C=180^@` [Sum of the co-interior angles]
`=>/_A+/_C=180^@`
`therefore ABCD` is a cyclic quadrilateral as the sum of the pair of opposite angles is `180^@`.