Question
In a $\triangle\text{ABC},$ $AB = BC = CA = 2a$ and $\text{AD}\perp\text{BC}.$ Prove that
$\text{AD}=\text{a}\sqrt{3}$
$\text{AD}=\text{a}\sqrt{3}$
✓
Answer
$\triangle\text{ABC},$ $AB = BC = AC = 2a$
$\triangle\text{AD}\perp\text{BC}$
AD bisects BC at D
$\text{BD}=\text{DC}=\frac{1}{2}\text{BC}=\text{a}$
Now in $\triangle\text{ABD}$
$AB^2 = AD^2 + BD^2$
$\Rightarrow (2a)^2 = AD^2 + a^2 $
$\Rightarrow 4a^2 = AD^2 + a^2$
$\Rightarrow AD^2 = 4a^2 - a^2 = 3a^2$ $=(\sqrt3\text{a})^2$
$\therefore\text{AD}=\sqrt{3}\text{a}$
$\triangle\text{AD}\perp\text{BC}$
AD bisects BC at D
$\text{BD}=\text{DC}=\frac{1}{2}\text{BC}=\text{a}$
Now in $\triangle\text{ABD}$
$AB^2 = AD^2 + BD^2$
$\Rightarrow (2a)^2 = AD^2 + a^2 $
$\Rightarrow 4a^2 = AD^2 + a^2$
$\Rightarrow AD^2 = 4a^2 - a^2 = 3a^2$ $=(\sqrt3\text{a})^2$
$\therefore\text{AD}=\sqrt{3}\text{a}$
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