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The options are:-

(A)equilateral (B)right angled (C)isosceles (D)either isosceles or right angled

Now I took examples to get to the answer but it was wrong. The answer is (D) but I got (C). To check for right angled, I took the values $3, 4, 5$ but the LHS and RHS were different, for equilateral i used $1,1,1$ again wrong. For isosceles I took angles to be $30^\circ, 30^\circ, 120^\circ$ and one of the same sides as $10$ but this was correct, so I want to know how is (D) correct?

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Another approach: Use the law of sines and a double angle identity:

\begin{align*} a\cos{A} &= b\cos{B} \\ \Rightarrow \frac{a\cos{A}}{\sin{A}} &= \frac{b\cos{B}}{\sin{A}} \\ \Rightarrow \frac{b\cos{A}}{\sin{B}} &= \frac{b\cos{B}}{\sin{A}} \\ \Rightarrow \cos{A}\sin{A} &= \cos{B}\sin{B} \\ \Rightarrow \sin{2A} &= \sin{2B} \end{align*}

Since $A$ and $B$ are both between 0 and 180 degrees, then either $2A = 2B$ or $2A = 180 - 2B$. The first case gives $A = B$, so the triangle is isosceles. The second case gives $A + B = 90$, so the third angle in the triangle is 90 degrees.

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By the law of cosines, one has $$\begin{align}a\cos A=b\cos B&\Rightarrow a\cdot\frac{b^2+c^2-a^2}{2bc}=b\cdot\frac{c^2+a^2-b^2}{2ca}\\&\Rightarrow a^2(b^2+c^2-a^2)=b^2(c^2+a^2-b^2)\\&\Rightarrow (a^2-b^2)(a^2+b^2-c^2)=0\end{align}$$

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  • $\begingroup$ yes, but what does it show? $\endgroup$ – dknight Apr 29 '15 at 18:09
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    $\begingroup$ @gaurav: We have $a^2=b^2$ or $a^2+b^2=c^2$. Can you take it from here? $\endgroup$ – mathlove Apr 29 '15 at 18:10
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    $\begingroup$ oh yes sorry thanks $\endgroup$ – dknight Apr 29 '15 at 18:14

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