A geometry problem by Vivek Vijayan

Let the midpoint of $BC$ be $M$ . Clearly, $AG=2GM$ since $G$ is the centroid. Also, $BM=CM=\dfrac{1}{2}AG\implies BM=CM=GM$

Thus, $\angle BGC=\boxed{90^{\circ}}$ .

Let $M$ denotes the midpoint of $BC$ .

It is by definition that $AGM$ is a straight line, and it follows that $AG=2GM$ .

Since $BM=MC$ , $AG=BC=BM+MC=2BM=2MC$ .

Therefore, $AG=2GM=2BM=2MC$ .

Therefore, $GM=BM=MC$ .

Therefore, a circle can be drawn with center $M$ and radii $GM$ , $BM$ and $MC$ , with $BC$ being the diameter.

Furthermore, $BGC$ is a semicircle.

Therefore, $\angle BGC=90^\circ$ .

Let $\displaystyle\,D$ be the mid point of $BC$ . Let $BD=DC=GD=x$ (say). Then we have $AG=2x$ .

Now $\angle\,GDC=\pi-\angle\,GDB$

$\rightarrow\,\cos\angle\,GDC=-\cos\angle\,GDB$

$\rightarrow\;\displaystyle\frac{x^2+x^2-GC^2}{2\cdot\,x\cdot\,x}=-\frac{x^2+x^2-BG^2}{2\cdot\,x\cdot\,x}$

Now simplification gives $BG^2+GC^2=4x^2=\left(2x\right)^2=BC^2\rightarrow\,\angle\,BGC=90^{\circ}$ .

It quite easy for me to find out its answer, which is 90 degree

Let O be the midpoint of BC. Now GO is a 1/2 AG. As AG is 2/3 AO. Also as O is midpoint BO = OC = 1/2 BC = 1/2 AG ( as AG=BC ). So GO = BO = OC. If you take BC as the diameter of a circle and O as the centre of the circle, now clearly CGB is right triangle, as G lies on the circumference. :)