Let a, B, C be the three sides of triangle ABC, and (C-B) x2 + 2 (B-A) x + A-B = 0, with two equal real roots, prove that triangle ABC is isosceles triangle

Let a, B, C be the three sides of triangle ABC, and (C-B) x2 + 2 (B-A) x + A-B = 0, with two equal real roots, prove that triangle ABC is isosceles triangle

Because there are two equal real roots, Δ = 0
Δ=4（b-a）^2-4*(c-b)*(a-b)
=4(b-a)*(b-a+c-b)
=4(b-a)(c-a)
=0
therefore
A = B or C = a
therefore
This triangle is isosceles Delta

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