# In ∆ABC, ∠C is an obtuse angle. AD ⊥ BC and AB2 = AC2 + 3 BC2

Question:

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 $\mathrm{BC}=\mathrm{CD}$.

Solution:

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

$\mathrm{AC}^{2}+3 \mathrm{BC}^{2}=\mathrm{AC}^{2}+\mathrm{BC}^{2}+2 \mathrm{BC} \times \mathrm{DC}$

$\Rightarrow 3 B C^{2}=B C^{2}+2 B C \times D C$

$\Rightarrow 2 \mathrm{BC}^{2}=2 \mathrm{BC} \times \mathrm{DC}$

$\Rightarrow B C=D C$