ABCD is a trapezium in which and AD = BC. Show that:
(i) $∠A = ∠BÂ$
(ii) $∠C = ∠D$
(iii) $∆ABC â‰\dots  ∆BAD$
(iv) Diagonal AC = Diagonal BD
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Given: ABCD is a trapezium in which AB ll CD and AD = BC.

To do: Here we have to show that:

(i) ∠A = ∠B 

(ii) ∠C = ∠D 

(iii) ∆ABC ≅ ∆BAD 

(iv) diagonal AC = diagonal BD

Solution:

Construction: Extend side AB. Draw a side CE || AD and CE = AD.

So, 

∠A $+$ ∠E = 180^o   

∠A = 180^o $-$ ∠E     ....(1)

Since,

AB || CD and AD || CE

Therefore, AECD is parallelogram.

Therefore, AD = CE

(i) ∠A = ∠B 

BC = CE (Given, AD = BC) 

Thus, in triangle BCE:

∠CBE = ∠E  (Angles opposite to equal sides of a triangle are equal)

So,

180^o $-$ ∠B = ∠E   (as, ∠CBE = 180^o $-$ ∠B)

∠B = 180^o $-$ ∠E     ....(2)

So, from (1) and (2):

∠A = ∠B

(ii) ∠C = ∠D

The measures of the adjacent angles of a parallelogram add up to be 180 degrees, or they are supplementary. So,

∠B $+$ ∠C = 180^o

And,

∠A $+$ ∠D = 180^o

As, ∠A = ∠B. So,

180^o $-$ ∠A = 180^o $-$ ∠B

∠D = ∠C

∠C = ∠D

(iii) ΔABC ≅ ΔBAD

BC = AD(given)

AB = BA(common)

∠B = ∠A (Proved above)

So, ΔABC ≡ ΔBAD by SAS criteria.  

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Updated on: 10-Oct-2022

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