Question

In the figure, CM is the perpendicular bisector of BD and $\angle ACB={47}^{\circ }$. Value of $\angle BCM$ is

• A. 66.5

Answer 1

The correct option is A 66.5
In $△BCM$ and $△CDM$,
$BM=MD$ [CM is the perpendicular bisector of BD]
$\angle CMB=\angle CMD={90}^{\circ }$ [CM is the perpendicular bisector of BD]
$CM=CM$ [Common side]

Hence, $△BCM$ and $△CDM$ are congruent.

So, $\angle BCM=\angle DCM$. [CPCT]

Now, $\angle ACD={180}^{\circ }\angle ACB+\angle BCM+\angle DCM={180}^{\circ }{47}^{\circ }+2\angle BCM={180}^{\circ }2\angle BCM={180}^{\circ }-{47}^{\circ }2\angle BCM={133}^{\circ }\angle BCM=\frac{{133}^{\circ }}{2}\angle BCM={66.5}^{\circ }$

Therefore, the value of $\angle BCM$ is ${66.5}^{\circ }$.