Question

In the figure, $AB=CD$ and $\angle A={90}^o=\angle D$. Then

• A. $\Delta ABC\cong \Delta DBC$ by SAS postulate
• B. $\Delta ABC\cong \Delta DCB$ by RHS postulate
• C. $\Delta ABC\cong \Delta DBC$ by AAS postulate
• D. $\Delta ABC\cong \Delta DCB$ by SSS postulate

Answer 1

Answer: (B)

$\Delta ABC\cong \Delta DCB$ by RHS postulate

In $\Delta ABC$ and $\Delta DBC$
$\angle A={90}^o=\angle D$ given
$BC=BC$ common side
$AB=CD$ given
$\therefore \Delta ABC\cong \Delta DCB$ (By RHS congruence postulate)