Question

ABCD is a trapezium in which AB II CD . If AD= BC show that angle A = Angle B and Angle C = Angle D

Answer 1

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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 )