In $\triangle \mathrm{ABC}, \angle \mathrm{C}$ is an obtuse angle. $\mathrm{AD} \perp \mathrm{BC}$ and $\mathrm{AB}^{2}=\mathrm{AC}^{2}+3 \mathrm{BC}^{2}$. Prove that $\mathrm{BC}=\mathrm{CD}$.
Given: $\triangle A B C$ where $\angle C$ is an obtuse angle, $A D \perp B C$ and $A B^{2}=A C^{2}+3 B C^{2}$
To prove: BC = CD
Proof:
In $\triangle \mathrm{ABC}, \angle \mathrm{C}$ is obtuse.
Therefore,
$\mathrm{AB}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}+2 \mathrm{BC} \times \mathrm{DC}$ (Obtuse angle theorem)......(1)
$\mathrm{AB}^{2}=\mathrm{AC}^{2}+3 \mathrm{BC}^{2}$ (Given) .....(2)
From (1) and (2), we get
$\mathrm{AC}^{2}+3 \mathrm{BC}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}+2 \mathrm{BC} \times \mathrm{DC}$
$\Rightarrow 3 B C^{2}=B C^{2}+2 B C \times D C$
$\Rightarrow 2 \mathrm{BC}^{2}=2 \mathrm{BC} \times \mathrm{DC}$
$\Rightarrow B C=D C$
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