Since D is the midpoint of the base, it divides the isosceles triangle into two congruent triangles, ABD and ACD
Because of symmetry of these triangles, we can let G and G1 will lie at (-a, b) and (a, b)
And the midpoint of GG! will lie on AD.....call this point, L
Suppose that AF and AG are the medians that intersect the base BC at M and N
And these medians pass through G and G1, respectively
Because GG1 are the same distance from the base BC, they are parallel to BC
So....triangles ADN and ALG1 are similar
This implies that AG1/LG1 = AN/DN
LG1 is the midpoint of GG1 = 2
And, because AN, is a median...then AG1 = (2/3)AN
Substituting, we have
(2/3)AN / 2 = AN / DN rearrange as
AN / (2/3)AN = DN /2
(3/2) = DN / 2
3 = DN
But this is (1/2) of DC...so DC = 6
And.....because of symmetry, DC = DB = 6
And because DN is perpendicular to BC.....then ADC is a right triangle
AD = 8, DC = 6......so.....AC is the hypotenuse = 10
And, by symmetry.....AC = AB = 10
So....the perimeter of ABC is
AC + AB + DB + DC =
10 + 10 + 6 + 6 =
32
