The medians AD, BE, and CF of triangle ABC intersect at the centroid G. The line through G that is parallel to BC intersects AB and AC at M and N , respectively. If the area of triangle AGM is 144, then find the area of triangle AGN.
Note that by property of centroid, AG : GD = 2 : 1.
Since MN is parallel to BC, MG is parallel to BD and GN is parallel to DC.
Since MG is parallel to BD, \(\triangle AMG \sim \triangle ABD\) by AA postulate.
Since GN is parallel to DC, \(\triangle AGN \sim \triangle ADC\) by AA postulate.
Then, \(MG = \dfrac23 BD = \dfrac23 DC = GN\).
Hence, G divides MN internally in the ratio 1 : 1.
Therefore, area of triangle AGN = area of triangle AGM = 144.