# Show that the DABC is an isosceles

Question:

Show that the DABC is an isosceles triangle if the determinant

$\Delta=\left[\begin{array}{ccc}1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B & \cos ^{2} C+\cos C\end{array}\right]=0$

Solution:

Given, $\quad \Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos ^{2} A+\cos A & \cos ^{2} B+\cos B & \cos ^{2} C+\cos C\end{array}\right|=0$

[Applying $C_{1} \rightarrow C_{1}-C_{3}$ and $C_{2} \rightarrow C_{2}-C_{3}$ ]

$\Rightarrow\left|\begin{array}{ccc}0 & 0 & 1 \\ \cos A-\cos C & \cos B-\cos C & 1+\cos C \\ \cos ^{2} A+\cos A-\cos ^{2} C-\cos C & \cos ^{2} B+\cos B-\cos ^{2} C-\cos C & \cos ^{2} C+\cos C\end{array}\right|=0$

Now,

[Taking ( $\cos A-\cos C$ ) common from $C_{1}$ and $(\cos B-\cos C)$ common from $C_{2}$ ]

$\Rightarrow$ $(\cos A-\cos C)(\cos B-\cos C) \times$

$\left|\begin{array}{ccc}0 & 0 & 1 \\ 1 & 1 & 1+\cos C \\ \cos A+\cos C+1 & \cos B+\cos C+1 & \cos ^{2} C+\cos C\end{array}\right|=0$

[Applying $C_{1} \rightarrow C_{1}-C_{2}$ ]

$\Rightarrow \quad(\cos A-\cos C)(\cos B-\cos C) \times$

$\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & 1+\cos C \\ \cos A-\cos B & \cos B+\cos C+1 & \cos ^{2} C+\cos C\end{array}\right|=0$

So,

$(\cos A-\cos C)(\cos B-\cos C)(\cos B-\cos A)=0$

$\cos A=\cos C$ or $\cos B=\cos C$ or $\cos B=\cos A$

$A=C$ or $B=C$ or $B=A$

Hence, $\triangle A B C$ is an isosceles triangle.