In △ABC, BM is a median, △BMC is equilateral and MC = 3cm. Through M is drawn line l∥AB that intersects CB at point N. Find:
- m∠ABC
- The distance from C to AB
- The distance from l to AB

In ABC BM is a median BMC is equilateral and MC 3cm Through M is drawn line lAB that intersects CB at point N Find mABC The distance from C to AB The distance f class=

Respuesta :

Answer:

m∠ABC = 60°

The distance from C to AB = 3 cm

The distance from l to AB = 1.5 cm

Step-by-step explanation:

The median of ΔABC = BM

The length of MC = 3 cm

Type of triangle given as ΔBMC  = Equilateral triangle

Line MN is parallel to AB and passes through M intersecting CB at N

Given that BM is a median, we have;

MC = AM = 3 cm

BM  = MC = CB = 3 cm, from ΔBMC  = Equilateral triangle

CN = NB by midpoint theorem

∴ CB = CN + NB = 2·CN = 3 cm

The distance from C to AB = CB = 3 cm

The distance from C to AB = 3 cm

CN = 3/2 = 1.5

CN = NB = 1.5

The distance from l to AB = CN = 1.5 cm

The distance from l to AB = 1.5 cm

m∠ABC = m∠BMC = m∠MBC = 60° Interior angles of an equilateral triangle.

m∠ABC = 60°