# Understand the problem

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# Start with hints

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Let \(a,b,c\) be the sides of a triangle and \(A,B,C\) be the angles opposite to those sides respectively. Given \( \sin (A-B)=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B- \cos A \sin B=\frac{a}{a+b}\sin A\cos B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\)

\(\Rightarrow \sin A\cos B-\frac{a}{a+b}\sin A\cos B=\cos A \sin B-\frac{b}{a+b} \cos A \sin B\) \(\Rightarrow \sin A\cos B(1-\frac{a}{a+b})=\cos A \sin B(1-\frac{b}{a+b})\) \(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\)

\(\Rightarrow \frac{b}{a+b}\sin A\cos B=\frac{a}{a+b}\cos A \sin B\) \(\Rightarrow b\sin A\cos B=a\cos A \sin B\) \(\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \)

\(\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{a}{\sin A} \) \(\Rightarrow (\frac{b}{\sin B})(\frac{\cos B}{\cos A})=\frac{b}{\sin B}\) [since \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)] \(\Rightarrow \frac{\cos B}{\cos A}=1\) \(\cos B=\cos A\) \(A=B\).

\(\Rightarrow \Delta ABC\) is isosceles (Proved).

# Connected Program at Cheenta

# I.S.I. & C.M.I. Entrance Program

Indian Statistical Institute and Chennai Mathematical Institute offer challenging bachelor’s program for gifted students. These courses are B.Stat and B.Math program in I.S.I., B.Sc. Math in C.M.I.

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