Question

ABCD is trapezium. AB || CD & AD=BC. prove angle A is equal to angle B​

Answer 1

Given: AB||CD and AD=BC

Construction:Draw perpendicular from A and B meeting DC at L and M respectively.

AB||CD and AL||BM(since sum of co-interior angles ALM and BML=180)

Hence, ABML is a parallelogram.

Furthermore, ABML is a rectangle.

Now in ΔADL and ΔBCM

AD=BC             (given)

AL=BM             (opposite sides of a rectangle)

∠ALD=∠BMC  (both 90°)

ΔADL≅ΔBCM (by RHS)

∠DAL=∠CBM

∠DAL+90°=∠CBM+90°

∠DAB=∠CBA

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 )