Question

In fig. $7.8,OA=OB$ and $OD=OC$. Show that (i) $\Delta AOD\cong \Delta BOC$ and (ii) $AD||BC.$

Answer 1

Given:- $OA=OBOD=OC$

To prove:- $\left(i\right)△AOD\cong △BOC\left(ii\right)AD\parallel BC$

Proof:-

Line $AB$ and $CD$ intersect.

$\angle AOD=\angle BOC\left(\text{Vertically opposite angle}\right)$

In $△AOD$ and $△BOC$,

$OA=OB\left(\text{Given}\right)$

$\angle AOD=\angle BOC$

$OD=OC\left(\text{Given}\right)$

$△AOD\cong △BOC\left(\text{SAS congruent rule}\right)$

$\angle OAD=\angle OBC$

But $\angle OAD$ and $\angle OBC$ and these form a pair of alternate angles if a transversal intersects two lines such that pair of alternate interior angles is equal, then lines are parallel.

$\therefore AD\parallel BC$