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Question
The following figure shows a trapezium ABCD in which AB is parallel to DC and AD = BC. 
Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.
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Solution
Given ABCD is a trapezium, AB || DC and AD = BC.
Prove that:
(i) ∠DAB = ∠CBA
(ii) ∠ADC = ∠BCD
(iii) AC = BD
(iv) OA = OB and OC = OD.
Proof: (i) Since AD || CB and transversal AC cuts them at A and O respectively.
Therefore, ∠A + ∠B = 180°
Since, AB || CD and AD || BC
Therefore, ABCD is a parallelogram.
∠A = ∠C
∠B = ∠D ....[ Since ABCD is a parallelogram ]
Therefore,
∠DAB = ∠CBA
∠ADC = ∠BCD
In ΔABC and ΔBAD, we have
BC = AD ....( given )
AB = BA ....( Common )
∠A = ∠B ....( proved )
ΔABC ≅ ΔBAD ....( SAS )
ΔABC ≅ ΔBAD
Since, Therefore AC = BD....( Corresponding parts of congruent triangles are equal. )
OA = OB
Again OC = OD ....( Since diagonals bisect each other at O )
Hence proved.
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