Question

ABCD is a trapezium with AB||CD and AD=BC. Prove that angleA=angleB

Answer 1

Draw BE//AD.
Therefore ABED is a Parallelogram
AD=BC (given) and
AD=BE(opposite sides of a parallelogram)
Therefore BC=BE
hence BCE is an isosceles triangle
therfore angleBCE=angleBEC ....(1)
angleADC + angleBEC= 180 (adjacent sides of a parallelogram is 180) ...(2)
angleBCD+angleBCE=180 (Linear pair) .....(3)
substituting  (1) in (3)
anldeBCD+anlgeBEC=180 ......(4)
from (1) and (4)
angle ADC=anlgeBCD
that is angle D=anlgeC ...(5)
angleA+angleD=180 (angles on the same side of transversal) ....(6)
angleB+angleC=180 ('' '') ...(7)
from (5) (6) and(7)
angle A= angle B
hence proved 

Answer 2

Given :- ABCD is a trapezium

AB || CD

AD = BC

To proof :-

(i)∠A = ∠B

(ii)∠C = ∠D

(iii)∆ ABC ≅ ∆ BAD

(iv)Diagonal AC = Diagonal = BD

Construction :- Draw DA || CE

Solution :-

(i) Since it's given ABCD is a trapezium

AB || CD

DA || CE ( By construction)

Therefore, ADCE is a parallelogram

So, DA = CE &

DC = AE ( Opposite side of parallelogram are equal )

But, AD = BC

Therefore, BC = CE ( Given )

∠CEB = ∠CBE ( In ∆ CBE angles opposite to equal sides are equal )

180° - ∠DAB = 180° - ∠ABC

[ ADCE is a parallelogram and ∠A + ∠E = 180° ∠B & ∠CBE form a linear pair ]

∠A = ∠B ( Cancelling 180° from both sides)

(ii) Co interior angles on the same side of a transversal are supplementary

∠A + ∠D = 180° & ∠B + ∠C = 180°

∠A + ∠D = ∠B + ∠C

∠B + ∠D = ∠B + ∠C ( ∠A = ∠B proved above)

∠D = ∠C

(iii) In ∆ ABC & ∆ BAD

AB = BA

∠B = ∠A ( proved above )

BC = BD ( Given )

∆ ABC ≅ ∆ BAD ( By SAS criteria)

(iv) AC = BD ( CPCT )