Prove that the opposite angles of an isosceles trapezium are supplementary?

Please help me with this question! :O

Dear Student!

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.

 

Cheers!

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Thank You sir! :D

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