Question

In the adjoining figure, $ABCD$ is a parallelogram and diagonals intersect at $O$. Find
(i)$\angle CAD$
(ii)$\angle ACD$
(iii)$\angle ADC$.

Answer 1

Here, $ABCD$ is a parallelogram.

$\angle CBD={46}^o$ [ Given ]

$AD\parallel BC$ and $AC$ is a transversal.

$\therefore$ $\angle CBD=\angle BDA$ [ Alternate angles ]

$\therefore$ $\angle BDA={46}^o$.

So, $\angle ODA={46}^o$

In $△ODA,$

$\Rightarrow$ $\angle ODA+\angle OAD+AOD={180}^o$

$\Rightarrow$ ${46}^o+\angle OAD+{68}^o={180}^o$

$\Rightarrow$ ${114}^o+\angle OAD={180}^o$

$\Rightarrow$ $\angle OAD={66}^o$

$\therefore$ $\angle CAD={66}^o$

$\Rightarrow$ $\angle D+\angle A={180}^o$ [ Sum of adjacent sides are supplementary in parallelogram ]

$\Rightarrow$ $(\angle CDB+\angle BDA)+(\angle CAD+\angle BAC)={180}^o$

$\Rightarrow$ $({30}^o+{46}^o)+({66}^o+\angle BAC)={180}^o$

$\Rightarrow$ ${142}^o+\angle BAC={180}^o$

$\Rightarrow$ $\angle BAC={38}^o$

$\Rightarrow$ $\angle BAC=\angle ACD$ [ Alternate angles ]

$\therefore$ $\angle ACD={38}^o$

Now,

$\Rightarrow$ $\angle ADC=\angle CDB+\angle BDA={30}^o+{46}^o={76}^o$

$\therefore$ We get, $\angle CAD={66}^o,\angle ACD={38}^o$ and $\angle ADC={76}^o$