Question

In the given figure ; $AD,BE$ and $CF$ are medians such that $AG=BC$ then $\angle BGC=$?

• A. ${90}^o$
• B. ${80}^o$
• C. ${60}^o$
• D. ${95}^o$

Answer 1

The correct option is A ${90}^o$

From the figure:

$\Rightarrow \angle GBC+\angle BGC+\angle GCB={180}^{\circ }$

$\Rightarrow \angle GBC+(\angle BGD+\angle CGD)+\angle GCB={180}^{\circ }$

$\Rightarrow (\angle GBC+\angle BGD)+(\angle CGD+\angle GCB)={180}^{\circ }$

$\Rightarrow 2\angle GBC+2\angle GCB={180}^{\circ }$

$\therefore \angle GBC+\angle GCB={90}^{\circ }$

$\therefore \angle BGC={180}^{\circ }-(\angle GBC+\angle GCB)$ by angle sum property of a triangle.

$\therefore \angle BGC={180}^{\circ }-{90}^{\circ }={90}^{\circ }$