Question
In an equilateral $\triangle\text{ABC},\ \text{AD}\perp\text{BC}$ prove that $AD^2 = 3BD^2.$
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Answer
Given: $\triangle\text{ABC}$ is an equilateral in which $AB = BC = CA.$
$\text{AD}\perp\text{BC}$
To prove: $AD^2 = 3BD^2$^
Proof: In equilateral $\triangle\text{ABC}.$
$\because\text{AD}\perp\text{BC}$
$\therefore$ AD bisects BC at D
$\therefore\text{BD}=\frac{1}{2}\text{BC}$
Now in right $\triangle\text{ABD}$
$AB^2 = AD^2 + BD^2$ (pythagoras theorem)
$\therefore AD^2 = AB^2 - BD^2$
$AD^2 = BC^2 - BD^2 {AB = AC = BC given}$
$AD^2 = (2BD)^2 - BD^2 {BC = 2BD}$
$AD^2 = 4BD^2 - BD^2$
$AD^2 = 3BD^2$^
Hence proved.
$\text{AD}\perp\text{BC}$
To prove: $AD^2 = 3BD^2$^
Proof: In equilateral $\triangle\text{ABC}.$
$\because\text{AD}\perp\text{BC}$
$\therefore$ AD bisects BC at D
$\therefore\text{BD}=\frac{1}{2}\text{BC}$
Now in right $\triangle\text{ABD}$
$AB^2 = AD^2 + BD^2$ (pythagoras theorem)
$\therefore AD^2 = AB^2 - BD^2$
$AD^2 = BC^2 - BD^2 {AB = AC = BC given}$
$AD^2 = (2BD)^2 - BD^2 {BC = 2BD}$
$AD^2 = 4BD^2 - BD^2$
$AD^2 = 3BD^2$^
Hence proved.
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