Question
✓
Answer
In $\triangle $AOD and $\triangle $BOC,
OA $ \times$ OB = OC $ \times$ OD
i.e $\frac{{OA}}{{OC}} = \frac{{OD}}{{OB}}$
And $\angle$ AOD = $\angle$ BOC [Vertically opposite Angles]
$\therefore $ $\triangle $AOD $ \sim $$\triangle $BOC [By SAS]
$\therefore $ $\angle$A = $\angle$C and $\angle$B = $\angle$D [Corresponding angles of similar $\triangle $ ]
OA $ \times$ OB = OC $ \times$ OD
i.e $\frac{{OA}}{{OC}} = \frac{{OD}}{{OB}}$
And $\angle$ AOD = $\angle$ BOC [Vertically opposite Angles]
$\therefore $ $\triangle $AOD $ \sim $$\triangle $BOC [By SAS]
$\therefore $ $\angle$A = $\angle$C and $\angle$B = $\angle$D [Corresponding angles of similar $\triangle $ ]
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