Question

Let $D$ be the middle point of the side $BC$ of a triangle $ABC$. If the triangle $ADC$ is equilateral, then $a^2:b^2:c^2$ is equal to

• A. $4:1:2$
• B. $4:1:3$
• C. $4:3:1$
• D. $3:4:1$

Answer 1

The correct option is B $4:1:3$


$\cos {120}^{\circ }=\frac{x^2+x^2-A{B}^2}{2x^2}\Rightarrow \frac{2x^2-A{B}^2}{2x^2}=\frac{-1}{2}\Rightarrow 3x^2=A{B}^2\Rightarrow AB=x\sqrt{3}\Rightarrow a^2:b^2:c^2=B{C}^2:A{C}^2:A{B}^2\Rightarrow a^2:b^2:c^2=(2x{)}^2:x^2:(x\sqrt{3}{)}^2=4:1:3$