Prove that the opposite angles of an isosceles trapezium are supplementary?
Please help me with this question! :O
Here is the answer to your query.
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.