Question

Let $ABC$ be a triangle and $D$ be the midpoint of $BC$. Suppose $\cot \left(\angle CAD\right):\cot \left(\angle BAD\right)=2:1$. If $G$ is the centroid of triangle $ABC$, then the measure of $\angle BGA$ is

• A. ${90}^{\circ }$
• B. ${105}^{\circ }$
• C. ${120}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is A ${90}^{\circ }$


Assuming $GD=a$ and $EG=k$
Then using the property of centroid
$AG=2a$
So the value of $AE=2a-k$
In $△BED$ and $△CFD$
We know thta $BD=CD$
$\angle EBD=\angle FCD$
Therefore the two triangles will be congurent.
$△BED=△CFD$
Given $\cot \left(\angle CAD\right):\cot \left(\angle BAD\right)=2:1\Rightarrow \cot x=2\cot y\Rightarrow \frac{AF}{CF}=2\frac{AE}{BE}\Rightarrow AF=2AE\Rightarrow AD+DF=2(2a-k)\Rightarrow 3a+DE=4a-2k\Rightarrow 3a+a+k=4a-2k\Rightarrow k=-2k\Rightarrow k=0$
So the point $E$ lies on the point $G$

Hence $\angle BGA={90}^{\circ }$