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