Question

ABCD is a parallelogram, A circle is passing through B, C, D and intersecting $\overline{AB}$ at E. If $\angle DAE={65}^{\circ }$ then find $\angle CDE$.

• A. ${65}^{\circ }$
• B. ${50}^{\circ }$
• C. ${55}^{\circ }$
• D. ${40}^{\circ }$

Answer 1

The correct option is A ${65}^{\circ }$

$\Rightarrow$ In given figure ABCD is a parallelogram passing through B, C, D and intersecting AB at E.

$\Rightarrow$ $\angle C=\angle A$ ---- - - ( 1 ) [Opposite angles of parallelogram are equal]

$\Rightarrow$ Since ABCD is a cyclic quadrilateral.

$\Rightarrow$ We know that "Exterior angle of cyclic quadrilateral is equal to interior opposite angle".

$\Rightarrow$ Hence $\angle C=\angle AED$ ----- ( 2 )

$\Rightarrow$ From ( 1 ) and ( 2 ), we get

$\Rightarrow$ $\angle A=\angle AED$ ---- ( 3 )

$\Rightarrow$ We know $\angle DAE={65}^{\circ }$

$\therefore$ $\angle AED={65}^{\circ }$ -----( 4 ) [From ( 3 )]

$\Rightarrow$ $\angle AED=\angle CDE$ --- ( 5 ) [Alternate angles between parallel lines]

$\therefore$ $\angle CDE={65}^{\circ }$ [From ( 4 ) and ( 5 )}