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Question
In \[∆ ABC, if \angle B = 60°,\] prove that \[\left( a + b + c \right) \left( a - b + c \right) = 3ca\]
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Solution
\[\text{ Given }, \angle B = 60°\]
\[\text{ We know that }, \cos B = \frac{a^2 + c^2 - b^2}{2ac}\]
\[ \Rightarrow \cos60° = \frac{a^2 + c^2 - b^2}{2ac}\]
\[ \Rightarrow \frac{1}{2} = \frac{a^2 + c^2 - b^2}{2ac} \left( \because \cos60° = \frac{1}{2} \right)\]
\[ \Rightarrow ac = a^2 + c^2 - b^2 \]
\[ \Rightarrow 3ac - 2ac = a^2 + c^2 - b^2 \]
\[ \Rightarrow 3ac = a^2 + c^2 - b^2 + 2ac\]
\[ \Rightarrow 3ac = a^2 + c^2 + 2ac - b^2 \]
\[ \Rightarrow 3ac = \left( a + c \right)^2 - b^2 \]
\[ \Rightarrow 3ac = \left( a + c + b \right)\left( a + c - b \right)\]
\[ \Rightarrow 3ac = \left( a + b + c \right)\left( a - b + c \right)\]
Hence proved.
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