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ABCD is a trapezium in which and AD = BC. Show that:
(i)
(ii)
(iii)
(iv) Diagonal AC = Diagonal BD
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 = 180o
∠A = 180o $-$ ∠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,
180o $-$ ∠B = ∠E (as, ∠CBE = 180o $-$ ∠B)
∠B = 180o $-$ ∠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 = 180o
And,
∠A $+$ ∠D = 180o
As, ∠A = ∠B. So,
180o $-$ ∠A = 180o $-$ ∠B
∠D = ∠C
∠C = ∠D
(iii) ΔABC ≅ ΔBAD
BC = AD(given)
AB = BA(common)
∠B = ∠A (Proved above)
So,