ABCD is a trapezium in which AB || CD and AD = BC (see figure): Show that
(i)      ∠A = ∠B
(ii)    ∠C = ∠D
(iii)    ∆ABC = ∆BAD
(iv)    diagonal AC = diagonal BD.



[Hint. Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Given: ABCD is a trapezium in which AB || CD and AD = BC.
To Prove: (i) ∠A = ∠B
(ii)    ∠C = ∠D
(iii)    ∆ABC ≅ ∆BAD
(iv)    diagonal AC = diagonal BD. Construction: Extend AB and draw a line
through C parallel to DA intersecting AB produced at E
Proof: (i) AB || CD    | Given
and    AD || EC    | By construction
∴ AECD is a parallelogram
| A quadrilateral is a parallelogram if a pair of opposite sides are parallel and are of
equal length
∴ AD = EC
| Opp. sides of a || gm are equal
But AD = BC    | Given
∴ EC = BC
∴ ∠CBE = ∠CEB    ...(1)
| Angles opposite to equal sides of a triangle are equal
∠B + ∠CBE = 180°    ...(2)
| Linear Pair Axiom
∵ AD || EC    | By construction
and transversal AE intersects them
∴ ∠A + ∠CEB = 180°    ...(3)
| The sum of consecutive interior angles on the same side of a transversal is 180°
From (2) and (3),
∠B + ∠CBE = ∠A + ∠CEB
But    ∠CBE = ∠CEB | From(1)
∴ ∠B = ∠A
or    ∠A = ∠B
(ii)    ∵    AB || CD
∠A + ∠D = 180°
| The sum of consecutive interior angles on a same side of a transversal is 180°
and    ∠B + ∠C = 180°
∴ ∠A + ∠D = ∠B + ∠C
But    ∠A = ∠B    | Proved in (i)
∴ ∠D = ∠C
or    ∠C = ∠D
(iii)    In ∆ABC and ∆BAD,
AB = BA    | Common
BC = AD    | Given
∠BC = ∠BAD    | From (i)
∴ ∆ABC ≅ ∆BAD.
| SAS Congruence Rule
(iv)    ∵ ∆ABC ≅ ∆BAD
| From (iii) above
∴ AC = BD. | C.P.C.T.