Let $ABC$ be a triangle and let $D$ be the mid-point of $BC$. Suppose $\cot \left(\angle CAD\right):\cot \left(\angle BAD\right)=2:1$. If $G$ is the centroid of $\mathrm{\Delta }ABC$, then the measure of $\angle BGA$ is

Let $\mathrm{\Delta }ABC$ and $D$ is the mid-point of side $BC$, 

then 

According to Apollonius theorem, 

$A{D}^2=\frac{1}{2}\left\{b^2+c^2-\frac{a^2}{4}\right\}$

$\Rightarrow AD=\frac{1}{2}\sqrt{2b^2+2c^2-a^2}$

$\because \frac{AG}{GD}=\frac{2}{1}$

$\Rightarrow AG=\frac{2}{3}AD=\frac{1}{3}\sqrt{2b^2+2c^2-a^2}$

Similarly $BG=\frac{1}{3}\sqrt{2c^2+2a^2-b^2}$

So, $\cos \theta =\frac{B{G}^2+A{G}^2-A{B}^2}{2\left(BG\right)\left(AG\right)}$, where $\theta =\angle AGB$

$=\frac{\frac{1}{9}\left(2c^2+2a^2-b^2\right)+\frac{1}{9}\left(2b^2+2c^2-a^2\right)-c^2}{2\left(BG\right)\left(AG\right)}$

$=\frac{a^2+b^2-5c^2}{18\left(BG\right)\left(AG\right)}$   $...\left(\mathrm{i}\right)$

Now, as $\cot {\theta }_2=2\cot {\theta }_1$

$\Rightarrow \frac{\cos {\theta }_2}{\sin {\theta }_2}=2\frac{\cos {\theta }_1}{\sin {\theta }_1}$

$\Rightarrow \frac{A{D}^2+b^2-\frac{a^2}{4}}{2AD·b\sin {\theta }_2}=2\frac{A{D}^2+c^2-\frac{a^2}{4}}{2AD·csin{\theta }_1}$

$\Rightarrow \frac{A{D}^2+b^2-\frac{a^2}{4}}{4\left(area\mathrm{\Delta }ADC\right)}=2\frac{A{D}^2+c^2-\frac{a^2}{4}}{4\left(area\mathrm{\Delta }ADB\right)}$

$\Rightarrow A{D}^2+b^2-\frac{a^2}{4}=2A{D}^2+2c^2-\frac{a^2}{2}$

$\left[\because \mathrm{\Delta }ADC=\mathrm{\Delta }ADB\right]$

$\Rightarrow A{D}^2=b^2+\frac{a^2}{4}-2c^2$

$\Rightarrow \frac{1}{4}\left(2b^2+2c^2-a^2\right)=b^2+\frac{a^2}{4}-2c^2$

$\Rightarrow 2b^2+2a^2=10c^2$

$\Rightarrow a^2+b^2=5c^2$  $...\left(\mathrm{ii}\right)$

From Eqs. $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we get 

$\cos \theta =0$

$\Rightarrow \theta =\frac{\pi }{2}$