In given figure ABCD is a trapezium in which AB∥CD and AD=BC. Show that (i)∠A=∠B (ii)∠C=∠D (iii)△ABC≅△BAD (iv) diagonal AC=diagonal BD

Given that ABCD is a trapezium, AB∥CD and AD=BC
To prove: (i)∠A=∠B (ii)∠C=∠D (iii)△ABC≅△BAD (iv)AC=BD
Construction: Draw CE∥AD and extend AB to intersect CE at E.
Poof:
(i)As AECD is a parallelogram. By construction, CE is parallel to AD and AE is parallel to CD, ∴AD=EC
But, AD=BC (Given)
∴BC=EC ⇒∠3=∠4 (Angle opposite to equal side are equal)
Now, ∠1+∠4=180° (consecutive interior angle of parallelogram )
And ∠2+∠3=180° (linear pair )
⇒∠1+∠4=∠2+∠3
⇒∠1=∠2 (∠3=∠4 so get cancelled with each other )
⇒∠A=∠B
(ii)AB∥CD ∴ ∠A+∠D=∠B+∠C=180° [Angles on the same side of transversal]
But, ∠A=∠B
Hence, ∠C=∠D
(iii)In △ABC and △BAD, we have
BC=AD (Given)
AB=BA (Common)
∠A=∠B (Proved)
Hence, by SAS congruence criterion, △ABC≅△BAD
(iv)We have proved that, △ABC≅△BAD
∴ AC=BD (CPCT)
Hence, all the four required results have been proved.