Question

In isosceles trapezium, prove that opposite angles are supplementary

Answer 1

Let ABCD be the trapezium with AD || BC and AB = DC.

Construction : Draw AE and DF perpendicular to BC.



Now in ΔABE and ΔDFC

AB = DC (Given)

AE = DF (Perpendicular distance between the parallel lines)

∠AEB = ∠DFC = 90°

∴ ΔABE ΔDCF (by RHS congurence criterion)

⇒ ∠ABE = ∠DCF ... (1)



Now AD || BC and BA is a transversal

⇒ ∠ABC + ∠BAD = 180°

⇒ ∠DCB + ∠BAD = 180° (from (1)) ......(2)



Also AD || BC and DC is a transversal

⇒∠ADC + ∠DCB = 180°

⇒∠ADC + ∠ABC = 180° (from (1)) ... (3)



from (2) and (3) we can conclude that the opposite angles of an isosceles trapezium are supplementary.



Cheers!