In the diagram below, D is the midpoint of the base of isosceles triangle ABC. Points G1 and G2 are the centroids of triangle ABC and triangle ACD, respectively. We know AD=8 and G1G2=4. What is the perimeter of triangle ABC?
From the diagram, BD = DC = G_1 G_2 = 4. Then from the Pythaogrean Theorem, AB = sqrt(4^2 + 8^2) = 4*sqrt(5), so the peimeter is AB + AC + BC = 4*sqrt(5) + 4*sqrt(5) + 8 = 8*sqrt(5) + 8.
+128408 Draw median AE to bisect BD
Let the intersection of G1G2 and AD = F
Since G1G2 is parallel to BC, then triangles A G1 F and AED are similar
But since AE is a median.....then AG1 is 2/3 of AE
So.... AE = 3/2 of AG1
So....ED is (3/2) of G1F = (3/2) (2) = 3
And, by symmetry, ED = (1/4) BC
So BC = 4ED = 4 * 3 = 12
So (1/2) BC = BD = 6
And by the Pythagorean Theorem
AD^2 + BD^2 = AB^2
8^2 + 6^2 = AB^2
100 = AB^2
10 = AB = AC
So....the perimeter of ABC = BC + AB + AC = 12 + 10 + 10 = 32
