Question

In the figure, $\angle A={90}^o=\angle D$ and $AB=DC$. Then $\Delta ABO\cong \Delta DCO$ by the congruence postulate

• A. ASA
• B. SAS
• C. SSS
• D. RHS

Answer 1

The correct option is A ASA

In $\Delta ABO$ and $\Delta DCO$
$\angle AOB=\angle COD$
$AB=CD$
$\angle BAO=\angle CDO={90}^o$
$\angle ABO={180}^o-(\angle BAO+\angle BOA)$
$={180}^o-(\angle COD+\angle COD)=\angle DCO$
$\therefore \Delta ABO\cong \Delta DCO$ (By ASA postulate)