Question

$M$ is a point on an arc $BC$ of a circle which circumscribes an equilateral $△ABC$. Then $AM$ is

• A. equal to $BM+CM$
• B. less than $BM+CM$
• C. greater than $BM+CM$
• D. equal to $|BM-MC|$

Answer 1

The correct option is A equal to $BM+CM$

Given,
$△ABC$ is an equilateral triangle.
$AB=BC=AC$
Extend $BM$ to point $D$. Such that $DM=CM$.
Since, $△ABC$ is an equilateral triangle.
$=>\angle A=\angle B=\angle C={60}^o$ and $AB=BC=AC$
Now, $ABCM$ is a cyclic quadrilateral.
$=>\angle BAC+\angle BMC={180}^o$
$\angle BMC={180}^o-{60}^o={120}^o$
Now, $\angle BMC+\angle CMD={180}^o$
$\angle CMD={180}^o-{120}^o={60}^o$
$\angle BAC=\angle CMD={60}^o$
Also, $\angle MDC=\angle MCD$ [Angles opposite to equal sides $CM$ and $DM$ are equal]
$△CMD$ is an equilateral triangle.
Now, In $△AMC$ and $BDC$, we have
$AC=BC$
$\angle CAM=\angle CBD$[Angles in the same segment of a circle are equal]
$\angle ABC=\angle AMC=\angle BDC={60}^o[\angle ABC$ and $\angle AMC$ lies in the same segment of a circle and are equal to each other.
$△AMC\cong △BDC$ (BY ASA)
$=>AM=BD$
$=>AM=BM+CM$ (As CM = DM)