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- Feb 14, 2012
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Inside a triangle $ABC$, there is a point P satisfies $\angle PAB=\angle PBC=\angle PCA=\lambda$. If the angles of the triangle are denoted by $\alpha$, $\beta$ and $\gamma$, prove that
$\dfrac{1}{\sin^2 \lambda}=\dfrac{1}{\sin^2 \alpha}+\dfrac{1}{\sin^2 \beta}+\dfrac{1}{\sin^2 \gamma}$
$\dfrac{1}{\sin^2 \lambda}=\dfrac{1}{\sin^2 \alpha}+\dfrac{1}{\sin^2 \beta}+\dfrac{1}{\sin^2 \gamma}$