Equilateral triangle ABC has a side length of 6 units. Point D lies on segment BC such that DC = 2(BD). What is the length of the altitude of triangle ADC from point C? Express your answer as a common fraction in simplest radical form.

Source: MathMovesU MATH CHALLENGE PROBLEM - 2012/04/12

Level: Junior

Solution 1.

Since 6 = BC = BD+DC = BD+2*BD = 3*BD, so BD = 6/3 = 2 and DC = 2*BD = 4. Let M be the midpoint of BC. Thus EC = 3 and DE = DC-EC = 1. Triangle ACE is a 30-60-90 triangle, so AE = AC*(sqrt(3)/2) = 3*sqrt(3). By the Pythagorean theorem, AD = sqrt((AE)^2+(DE)^2) = sqrt(28) = 2*sqrt(7). Let h is the the length of the altitude of triangle ADC from point C. The area of triangle ADC can be computed in two different ways, namely (AD)h/2 = sqrt(7)*h and (DC)(AE)/2 = 4*3*sqrt(3)/2 = 6*sqrt(3). Equating the two expressions and solve for h,
sqrt(7)*h = 6*sqrt(3),
h = 6*sqrt(21)/7.

Solution 2.

Since 6 = BC = BD+DC = BD+2*BD = 3*BD, so BD = 6/3 = 2 and DC = 2*BD = 4. By the law of cosines,
(AD)^2
= (AC)^2+(DC)^2-2(AC)(DC)cos(C),
= 6^2+4^2-2(6)(4)cos(60)
= 28.
So AD = 2*sqrt(7). Let h is the the length of the altitude of triangle ADC from point C. The area of triangle ADC can be computed in two different ways, namely (AD)h/2 = sqrt(7)*h and (AC)(DC)sin(C)/2 = (AC)(DC)sin(C)/2 = 6*4*sin(60) = 24*sqrt(3)/2 = 6*sqrt(3). Equating the two expressions and solve for h,
sqrt(7)*h = 6*sqrt(3),
h = 6*sqrt(21)/7.