Question

In the adjoining figure $AB\parallel CD$, prove that $\Delta AOB\sim \Delta DOC$

Answer 1

$AB\parallel CD$ ( given )

In $△AOB$ and $△DOC$,

$\angle AOB=\angle DOC$ (Vertically opposite angles)

$\angle OAB=\angle ODC$ (Alternate angles)

$\angle ABO=\angle DCO$ (Alternate angles)

From $AAA$ (angle-angle-angle) similarity theorem,

$△AOB\sim △DOC$