Given:- ABCD is a trapezium where AB∥CD and AD=BC
To prove:- AC=BD
Construction:- Extend AB and draw a line through C parallel to AD intersecting AB produced at E.
Proof:-
AD∥CE(From construction)
AE∥CD(As AB∥CD,&AB produced at E)
In quadrilateral AECD, both the pair of opposite sides are parallel.
∴AECD is a parallelogram.
∴AD=CE.....(1)(∵Opposite sides of a parallelogram are equal)
AD=BC.....(2)(Given)
From equation (1)&(2), we have
∴BC=CE
⇒∠CEB=∠CBE.....(3)(∵Angle opposite to equal sides are equal)
Now, for AD∥CE and AE is transversal,
∠A+∠CEB=180°
⇒∠A=180°−∠CEB.....(4)
Also AE is a line,
∠B+∠CBE=180°(Linear pair) { ∠CBA=∠B }
⇒∠B+∠CEB=180°(From (3))
⇒∠B=180°−∠CEB.....(5)
Now, from equation (4)&(5), we get
∠A=∠B.....(6)
In △ABC and △BAD
AB=BA(Common)
∠A=∠B(From (6))
BC=AD(Given)
By SAS congruet rule,
△ABC≅△BAD
Therefore, by C.P.C.T.,
AC=BD
Hence proved.