In a triangle, the three medians (joining vertex to midpoint of opposite side) intersect at a point called the centroid, usually denoted as \( G \).
That is,
\[ \frac{AG}{GD} = \frac{BG}{GE} = \frac{CG}{GF} = \frac{2}{1} \]
Q. \( \triangle ABC \) is a triangle and \( G \) is its centroid. If \( AG = BG \), what is \( \angle BGC \)?
We know from centroid property that \( AG:GD = 2:1 \), but here we are told \( AG = BG \).
This implies triangle symmetry such that:
- \( AG = BG \)
- \( D \) is the midpoint of \( BC \)
- \( AG \perp BC \)
\[ \Rightarrow \angle BGC = 90^\circ \]
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