MCQ
If $\triangle\text{ABC}$ is an equilateral triangle such that $\text{AD}\perp\text{BC},$ then $AD^2 =$
- A$\frac{3}{2}\text{DC}^2$
- B$\text{2DC}^2$
- ✓$3\text{CD}^2$
- D$4\text{DC}^2$
✓
Answer
Correct option: C.
$3\text{CD}^2$
In equilateral $\triangle\text{ABC},\ \text{AD}\perp\text{BC} , AD$ bisects $BC$ at $D$
$\therefore BD = DC$
Now in right $\triangle\text{ADC},$
$AC^2 = AD^2 + DC^2 ($Pythagoras Theorem$)$
$AD^2 = AC^2 - DC^2$
$= BC^2 - DC^2 ( \because AC = BC = AB)$
$= (2DC)^2 - DC^2 ( \because D$ is mid point of $BC)$
$= 4DC^2 - DC^2 = 3DC^2$
$= 3CD^2$
$\therefore BD = DC$
Now in right $\triangle\text{ADC},$
$AC^2 = AD^2 + DC^2 ($Pythagoras Theorem$)$
$AD^2 = AC^2 - DC^2$
$= BC^2 - DC^2 ( \because AC = BC = AB)$
$= (2DC)^2 - DC^2 ( \because D$ is mid point of $BC)$
$= 4DC^2 - DC^2 = 3DC^2$
$= 3CD^2$
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