Congruency, SMB-006

In the quadrilateral \(ABCD\), \(AB \parallel CD, \angle BAD = \angle BCD\) and \(\angle DBC = \angle BDA = 90^{\circ} \).

Prove that this pair of triangles are congruent. (2 marks)

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\(\angle BAD = \angle BCD\ \ \text{(given)} \)

\(\angle DBC = \angle BDA = 90^{\circ} \ \ \text{(given)} \)

\(BD\ \text{is common} \)

\(\therefore\ \Delta BAD \equiv \Delta DCB\ \ \text{(AAS)}\)