Question

In the following figure, OA =OC and AB =BC.

Prove that :

(i) $\angle$ AOB = ${90}^0$

(ii) $\Delta$ AOD $\cong \Delta$ COD

(iii) AD =CD

Answer 1

Proof,
In $△$ ABO and $\Delta$ CBO
AB=CB
BO=BO {commom among $\Delta$ ABO and $\Delta$ CBO
AO=CO
So, $\Delta ABO\cong \Delta CBO$ by SSS
$\angle ABO=\angle CBO$ {cpctc }
$\angle AOB=\angle COB$ {cpctc}
Let AOB=COB=X
X+X= ${180}^0$
2X=${180}^0$
X=${90}^0$

$\angle AOB={90}^0$ Proved .

In $\Delta$ ABD and $\Delta$ CBD
​AB=CB
ABO=CBO
BD {COMMON}So $\Delta ABD\cong \Delta$ CBD by SAS

AD = CD { cpctc}

​PROVED