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 $\triangle\text{ADC},$
$A C^2=A D^2+D C^2 \ ($Pythagoras Theorem$)$
$A D^2=A C^2-D C^2$
$=B C^2-D C^2\ (\because AC = BC = AB)$
$=(2 D C)^2-D C^2\ (\because D$ is mid point of $BC)$
$=4 D C^2-D C^2=3 D C^2$
$=3 C D^2$
$AD$ bisects $BC$ at $D$
$\therefore BD = DC$
Now in right $\triangle\text{ADC},$
$A C^2=A D^2+D C^2 \ ($Pythagoras Theorem$)$
$A D^2=A C^2-D C^2$
$=B C^2-D C^2\ (\because AC = BC = AB)$
$=(2 D C)^2-D C^2\ (\because D$ is mid point of $BC)$
$=4 D C^2-D C^2=3 D C^2$
$=3 C D^2$
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