Question

AD = AC
If the above statement is true then mention answer as 1, else mention 0 if false

Answer 1

In $△AOB$ and $△OBC$,
$OA=OC$ (Given)
$AB=BC$ (Given)
$OB=OB$ (COmmon)
Thus, $△AOB\cong △COB$ (SSS rule)
Hence, $\angle AOB=\angle COB=x$ (By cpct)
We know, $\angle AOB+\angle COB=180$ (Angles on a straight line)
$x+x=180$
$x={90}^{\circ }$
Thus, $\angle AOB=\angle COB={90}^{\circ }$
$\angle AOB=\angle =COD={90}^{\circ }$ (Vertically opposite angles) (I)
$\angle COB=\angle =AOD={90}^{\circ }$ (Vertically opposite angles) (II)
Thus, from I and II
$\angle COD=\angle AOD$ ..(III)
Now, In $△AOD$ and $△COD$,
$AO=OC$ (Given)
$\angle AOD=\angle COD$ (From III)
$OD=OD$ (Common)
thus, $△AOD\cong △COD$ (SAS rule)
Hence, $AD=CD$ (By cpct)