In Figure, `A B C D` is a trapezium in which `A B C D` and `A D=B Cdot` show that : `/_A=/_B` (ii) `/_C=/_D` ` A B C~= B A D` diagonal `A C=d i agon a lB D`

Explanation

Given ABCD is trapezium where `AD=BC`
To prove, `/_A=/_B`
we can see that AECD is a parallelogram, so sum of adjacent angles`=180^0`
`/_A+/_E=180^0
/_A+x=180^0
/_A=180^0−x=/_B`
Hence proved.
(ii) To prove,`/_C=/_D`,
sum of adjacent angles in parallelogram is π
so,`/_D/_C+180^0−2x=180^0`
`/_C+/_D=2x``180^0−x+/_C=180^0`
`/_C=x,/_D=x`
`/_C=/_D` Hence proved.
(iii)`/_\ABC=/_\BAD`side AB is common.
`AD=BC`(given)
so the angle including both the sides is also same.
`/_A=/_B`
`/_\ABC=/_\BAD`(By SAS congruent Rule)
Hence proved.
(iv) As `/_\ABC=/_\BAD`
The third side of both triangles i.e. diagonals are equal `AC=BD`