Question

In the given figure $AB$ is diameter of circle with centre $O$ and chord $ED$ is parallel to $AB$ and $\angle EAB={65}^o$. Then $m\angle EBD$ is:

• A. ${55}^o$
• B. ${65}^o$
• C. ${25}^o$
• D. ${40}^o$

Answer 1

Answer: (D)

${40}^o$

Since $AB$ is diameter, $\angle AEB={90}^{\circ }$.

In $△AEB$,

$\angle ABE+\angle EAB+\angle AEB={180}^o$ ...[Angle sum property]

$⟹$$\angle ABE={180}^o-\angle EAB-\angle AEB$

$⟹$$\angle ABE={180}^o–{65}^o-{90}^o={25}^{\circ }$.

Given, $ED\parallel AB$,

$⟹$ $\angle DEB=\angle EBA={25}^{\circ }$ ....[Alternate interior angles].

Since $EDCB$ is a cyclic quadrilateral,

$\angle EAB+\angle EDB={180}^{\circ }$ ...[Opposite angles of cyclic quadrilateral are supplementary]

$⟹$$\angle EDB={180}^o–{65}^o={115}^{\circ }$.

In $△EDB$,

$\angle EBD+\angle DEB+\angle BDE={180}^o$

$⟹$$\angle EBD=180-\angle DEB-\angle BDE$

$⟹$$\angle ABE=180–25-115={40}^{\circ }$.

Hence, option $D$ is correct.