In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(i) BD = AC.
(ii) ΔBCD ≅ ΔADC

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Solution

The given parameters from the questions are ∠DAB = ∠CBA and AD = BC.
(i) ΔABD and ΔBAC are similar by SAS congruence as
AB = BA (It is the common arm)
∠DAB = ∠CBA and AD = BC (These are given in the question)
So, triangles ABD and BAC are similar i.e. ΔABD ≅ ΔBAC. (Hence proved).
It is now known that ΔABD ≅ ΔBAC so,
BD = AC (by the rule of CPCT).
(ii) Now, ΔBCD and ΔADC are similar by SSS congruence as
CD = DC (It is the common arm)
BD = AC (from (i))
and AD = BC (These are given in the question)
So, ΔBCD ≅ ΔADC (hence proved)

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In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that (i) BD = AC. (ii) ΔBCD ≅ ΔADC

In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that
(i) BD = AC.
(ii) ΔBCD ≅ ΔADC