Question

In the given figure, $AB=BC=CD$. If $\angle BAC={25}^{\circ }$, then value of $\angle AED$ is:

• A. ${50}^{\circ }$
• B. ${60}^{\circ }$
• C. ${65}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

Answer: (D)

${75}^{\circ }$

In $△ABC$, $AB=AC$
Thus, $\angle BAC=\angle BCA$ ....(Isosceles triangle property).
Also, $\angle BDC=\angle BAC={25}^{\circ }$ ....(Angles in same segment).

In $△BDC$,
by angle sum property, the sum of angles $={180}^o$
$⟹$ $\angle CBD+\angle BCA+\angle ACD+\angle BDC=180$
$⟹$ $\angle ACD=180-75={105}^{\circ }$.
Now, in cyclic quadrilateral $ACDE$,
$\angle ACD+\angle AED=180$ ....(Opposite angles of cyclic quadrilateral)
$⟹$ $\angle AED=180-105$
$⟹$ $\angle AED={75}^{\circ }$

Hence, option $D$ is correct.