The angles of triangle ABC are pi/7, 2pi/7 and 4pi/7.  For all these angles tan(7x) = 0.

Let T = tan(x), then the multiangle formula for tan(7x) can be written as (7T - 35T^3 + 21T^5 - T^7) / (1 - 21T^2 + 35T^4 - 7T^6).  

See http://mathworld.wolfram.com/Tangent.html and http://oeis.org/A034839

Since we are looking for tan(7x) = 0 then the seven roots of 7T - 35T^3 + 21T^5 - T^7 = 0 correspond to the values of tan(n*pi/7) for n=0 to 6.

This can be factored over sqrt(7) to get T * (-1)*(-sqrt(7) - 7T - sqrt(7)T^2 + T^3) * (sqrt(7) - 7T + sqrt(7)T^2 + T^3) = 0. 
T = tan(0) is obvious; the other six tan(n*pi/7) values are split among the two cubics.

Basic identities can be used to show -tan(pi/7) = tan(6pi/7), -tan(2pi/7) = tan(5pi/7), and -tan(3pi/7) = tan(4pi/7). 
If any of these pairs were roots of the same cubic then the cubic would have a form of a*b + a*T + bT^2 + T^3 for some values a and b. 
This is not the case, so each cubic has to have one root from each pair.

Neither cubic has all same sign or strictly alternating signs, so each cubic has at least one positive and one negative root.
Looking at the constant terms of the cubics, the first cubic has two negative roots and a postive root and the second cubic is the complement with two positive roots and a negative root. 
Then looking at the linear terms of the cubics, the root of opposite sign is the largest root by absolute value.

Basic tan identities can be used to create tan(4pi/7) < (5pi/7) < tan(6pi/7) < 0 tan(pi/7) < tan(2pi/7) < tan(3pi/7). 
Then to satisfy the observations about the cubics, the roots must be tan(5pi/7), tan(6pi/7) and tan(3pi/7) for -sqrt(7) - 7T - sqrt(7)T^2 + T^3 = 0 and tan(4pi/7), tan(pi/7) and tan(2pi/7) for sqrt(7) - 7T + sqrt(7)T^2 + T^3 = 0.

So finally, the quadratic coefficient of sqrt(7) - 7T + sqrt(7)T^2 + T^3 = 0 immediately implies tan(4pi/7) + tan(pi/7) + tan(2pi/7) = -sqrt(7).