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In Figure, A B C D is a trapezium in whi...

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`

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Given ABCD is trapezium where AD=BC.
(i) To prove: `/_A=/_B`
we can see that AECD is a parallelogram, so sum of adjacent angles `=180^@`
`/_A+/_E=180^@`
`/_A+x=180^@`
`/_A=180^@−x=/_B`
Hence proved.
(ii) To prove: `/_C=/_D`
sum of adjacent angles in parallelogram is `pi`, so
`/_D/_C+180^@−2x=180^@`
`/_C+/_D=2x`
Now
`/_B+/_C=180^@`
`180^@−x+/_C=180^@=0` `/_C=x`, so
`/_D=x`And `/_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`. So
`/_\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
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