Question

In the following figure, $OA=OC$ and $AB=BC$. Prove that :
i) $\angle AOB={90}^o$
ii) $\Delta AOD\cong \Delta COD$
iii) $AD=CD$

Answer 1

i)

$\angle AOB=\angle COB\left(CPCT\right)$

Let

$\angle AOB=\angle COB=x$

x+x=180

2x=180

x=90

$\therefore \angle AOB=90$

ii)

AO=OC

AD=CD

OD=OD

$\therefore △AOD\cong △COD$

iii)

In $△ABD,△CBD$

AB=CB

ABO=CBO

BD=BD

$△ABD\cong △CBD$

$\therefore AD=CD\left(CPCT\right)$