# Triangle ABC is regular. Point O is its center. Line OM is perpendicular to plane ABC

**Triangle ABC is regular. Point O is its center. Line OM is perpendicular to plane ABC. Prove that MA = MB = MC. Find MA if AB = 6cm, MO = 2cm.**

Since the triangle ABC is correct, then AC = AB = BC = 6 cm.

The center of the triangle is the point of intersection of its heights, and since the triangle ABC is regular, its medians and bisectors are also.

Then the length of the segment AH = BC = AB / 2 = 6/2 = 3 cm, and the triangle ACH is rectangular.

By the Pythagorean theorem, CH ^ 2 = AC ^ 2 – AH ^ 2 = 36 – 9 = 27.

CH = √27 = 3 * √3 cm.

The medians of the triangle, at the point of their intersection, are divided in a ratio of 2/1 starting from the apex.

OH = OC / 2.

OC + (OC / 2) = CH = 3 * √3 cm.

3 * OC = 6 * √3.

ОC = 2 * √3 cm.

ОА = ОC = ОВ = 2 * √3 cm.

The right-angled triangles AOM, BOM, COM are equal in two legs, and then AM = BM = CM, which was required to prove.

The AOM triangle is rectangular, in which, according to the Pythagorean theorem, AM ^ 2 = AO ^ 2 * OM ^ 2 = 12 + 4 = 16.

AM = √16 = 4 cm.

Answer: The length of the segment AM = 4 cm.