ABCD is a trapezium in which AB || CD and AD = BC (see the given figure). From the figure, which of the following can be deduced?

$Δ A B C ≅ Δ B C D$

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$Δ A B C ≅ Δ B C E$

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$Δ A B C ≅ Δ B A D$

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$N o n e o f t h e s e$

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Solution

The correct option is C
$Δ A B C ≅ Δ B A D$

ABCD is an isosceles trapezium (given)
Hence, $∠ D A B = ∠ C B A$ ...(base angles of an isosceles trapezium)
In $△ B A D$ and $△ A B C$ ,
$(i) A D = B C$ ...(given)
$(i i) ∠ D A B = ∠ C B A$ ... (base angles of an isosceles trapezium)
$(i i i) A B = A B$ ...(common)
$(i v) △ ≅ △ A B C$ ...(SAS Postulate)

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