Question

$ABCD$ is a parallelogram. A circle passes through $A$ and $D$ and cuts $AB$ at $E$ and $DC$ at $F.$ Given $\angle BEF={80}^{\circ },$ find $\angle ABC.$

• A. ${100}^{\circ }$
• B. ${40}^{\circ }$
• C. ${80}^{\circ }$
• D. ${104}^{\circ }$

Answer 1

The correct option is C ${80}^{\circ }$

Given, $ABCD$ is a parallelogram and $AEFD$ is a cyclic quadrilateral.
$\angle BEF=80\circ$
Now, $\angle ADC=\angle BEF={80}^{\circ }$ (Angle of a cyclic quadrilateral is equal to the opposite exterior angle )
Also, now in parallelogram ABCD,
$\angle ABC=\angle ADC={80}^{\circ }$ (Opposite angles of a parallelogram are equal)