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# If for a triangle abc tan a-b plus tan b-c plus tan c-a equals 0 then what can you say about the triangle?

tan (A-B) + tan (B-C) + tan (C-A)=0

tan (A-B) + tan (B-C) - tan (A-C)=0

tan (A-B) + tan (B-C) = tan (A-C)

(A-B) + (B-C) = A-C

So we can solve

tan (A-B) + tan (B-C) = tan (A-C)

by first solving

tan x + tan y = tan (x+y)

and then substituting x = A-B and y = B-C.

tan (x+y) = (tan x + tan y)/(1 - tan x tan y)

So tan x + tan y = (tan x + tan y)/(1 - tan x tan y)

(tan x + tan y)tan x tan y = 0

So, tan x = 0 or tan y = 0 or tan x = - tan y

tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = - tan(B-C)

tan(A-B) = 0 or tan(B-C) = 0 or tan(A-B) = tan(C-B)

A, B and C are all angles of a triangle, so are all in the range (0, pi).

So A-B and B-C are in the range (- pi, pi).

At this point I sketched a graph of y = tan x (- pi < x < pi)

By inspection I can see that:

A-B = 0 or B-C = 0 or A-B = C-B or A-B = C-B +/- pi

A = B or B = C or A = C or A = C +/- pi

But A and C are both in the range (0, pi) so A = C +/- pi has no solution

So A = B or B = C or A = C

A triangle ABC has the property that tan (A-B) + tan (B-C) + tan (C-A)=0 if and only if it is isosceles (or equilateral).

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