Question

In figure, seg $AB$ is a diameter of a circle with centre $O.$ The bisector of $\angle ACB$ intersects the circle at point $D.$
then $\mathrm{seg}AD\cong \mathrm{seg}BD.$

• A. True
• B. False

Answer 1

The correct option is A True

Draw seg $OD.$

$\Rightarrow$ $\angle ACB={90}^o$ [ Angle inscribed in semicircle ]

$\Rightarrow$ $\angle ACB={45}^o$ [ $CD$ is the bisector of $\angle C$ ]

$\Rightarrow$ $m\left(arcDB\right)=2\angle ACB={90}^o$ [ Inscribed angle theorem ]

$\Rightarrow$ $\angle DOB={90}^o$ [ Definition measure of an arc ] ----- ( 1 )

$\Rightarrow$ $segOA\cong segOB$ [ Radii of the circle ] ----- ( 2 )

$\therefore$ Line $OD$ is perpendicular bisector of $segAB$ [ From ( 1 ) and ( 2 ) ]

$\therefore$ $segAD\cong segBD$