Question

Let ABC be an equilateral triangle inscribed in circle O. M is a point on arc BC. Lines AM, BM and CM are drawn. Then $\overline{AM}$ is:

• A. equal to $\overline{BM}+\overline{CM}$
• B. Less than $\overline{BM}+\overline{CM}$
• C. greater than $\overline{BM}+\overline{CM}$
• D. equal, less than, or greater than $\overline{BM}+\overline{CM}$, depending upon the position upon M

Answer 1

The correct option is A equal to $\overline{BM}+\overline{CM}$

Let us consider ideal case.

O is the center which lies on AM.

$\therefore$ AM=2R, R is the radius of the circle.

AM$\perp$BC

O be a point on AM and BC

$\therefore$AD$\perp$BC

$\frac{BD}{OB}=\frac{\sqrt{3}}{2}$

Let AB = BC = CA = a, OA = OB = OC = R

$\frac{a}{2R}=\frac{\sqrt{3}}{2}$

$a=\sqrt{3}R$

AM = 2R, BM = CM = $\frac{BD}{\cos 30°}=\frac{a}{\sqrt{3}}$

BM+CM = $\frac{2a}{\sqrt{3}}$ = 2R = AM

$\therefore \overline{AM}=\overline{BM}+\overline{CM}$