If G is the centroid of ΔABC and AG = BC, then ∠BGC is:
AG = 2GM
As AG = BC for the given condition
2GM = BC (where M is the midpoint of BC)
thus BM = MG = MC
∠ MGB and ∠GBM are equal
∠MGC and ∠GCM are equal
so ∠MGB + ∠MGC = ∠GBM + ∠GCM
that is ∠BGC = ∠GBM + ∠GCM
As the sum of all three angles is 180°
∠BGC is 90°
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