Question

Prove that in the following kite, $\Delta ADC$ is congruent to $\Delta ABC$. Given that AD = AB and $\angle ADC$ = $\angle ABC$ = ${90}^{\circ }$
[2 MARKS]

Answer 1

Steps: 1 Mark
Proof: 1 Mark

In $\Delta ADCand\Delta ABC$

$AD=AB$ [Given]

$\angle ADC=\angle ABC={90}^o$ [Given]

$AC=CA$ [common]

Hence, $\Delta ADC\cong \Delta ABC$ [By RHS congruence condition]