Question

If $l$ is the length of the median from the vertex $A$ to the side $BC$ of a $△ABC$, then:

• A. $4l^2=2b^2+2c^2-a^2$
• B. $4l^2=b^2+c^2+2bc\cos A$
• C. $4l^2=a^2+4bc\cos A$
• D. $4l^2={(2s-a)}^2-4bc{\sin }^2\left(\frac{A}{2}\right)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $4l^2=2b^2+2c^2-a^2$
B $4l^2=b^2+c^2+2bc\cos A$
C $4l^2=a^2+4bc\cos A$

$\because D$ is the mid-point
$\therefore {AB}^2+{AC}^2=2[{AD}^2+{BD}^2]$
$\Rightarrow c^2+b^2=2[l^2+{\left(\frac{a}{2}\right)}^2]$
$\Rightarrow 2c^2+2b^2=4l^2+a^2$
$\Rightarrow 4l^2=2b^2+2c^2-a^2$
$\Rightarrow 4l^2=b^2+c^2+(b^2+c^2-a^2)$
$\Rightarrow 4l^2=b^2+c^2+2bc\cos A$ (using cosine rule)
$\Rightarrow 4l^2=(b^2+c^2-a^2)+a^2+2bc\cos A$
$\Rightarrow 4l^2=2bc\cos A+a^2+2bc\cos A$
$\Rightarrow 4l^2=4bc\cos A+a^2$