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 $\triangle\text{ABD}$
$ A B^2=A D^2+B D^2$
$ \Rightarrow(2 a)^2=A D^2+a^2 \Rightarrow 4 a^2=A D^2+a^2$
$ \Rightarrow A D^2=4 a^2-a^2=3 a^2=(\sqrt{3} 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 $\triangle\text{ABD}$
$ A B^2=A D^2+B D^2$
$ \Rightarrow(2 a)^2=A D^2+a^2 \Rightarrow 4 a^2=A D^2+a^2$
$ \Rightarrow A D^2=4 a^2-a^2=3 a^2=(\sqrt{3} a)^2$
$\therefore\text{AD}=\sqrt{3}\text{a}$
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