91Ó°ÊÓ

$\stackrel{¯}{AD}$ and $\stackrel{¯}{BE}$ are congruent medians of $â–\mu ABC$.

Basic concept of geometry related to triangles.

GD=$\frac{1}{3}$AD, and

GE= $\frac{1}{3}$BE.

Given that $\stackrel{¯}{AD}$and $\stackrel{¯}{BE}$are congruent-medians of $\mathrm{Δ}ABC$.

Hence, AD=BE.

Multiply both sides of equation with $\frac{1}{3}$, one gets,

$\frac{1}{3}$AD= $\frac{1}{3}$BE.

Hence, GD=GE.

GA= $\frac{2}{3}$AD, and

GB=$\frac{2}{3}$BE.

Given that $\stackrel{¯}{AD}$and $\stackrel{¯}{BE}$are congruent-medians of $\mathrm{Δ}ABC$.

Hence, AD=BE.

Multiply both sides of equation with $\frac{2}{3}$, one gets

$\frac{2}{3}$AD= $\frac{2}{3}$BE.

Hence, GA=GB.

c.Consider the triangle GAB,

Since the two sides of the triangle GA and GB are equal.

$\mathrm{Δ}GAB$is an isosceles triangle.

Thus, $∠GAB=∠GBA$.

Consider the triangle GED,

Since the two sides of the triangle are equal, that is, GE=GD. $\mathrm{Δ}GED$is an isosceles triangle.

Thus $∠GED$= $∠GDE$.

From the figure, we have $\stackrel{¯}{ED}$parallel to$\stackrel{¯}{AB}$.

Hence, the alternate angles are also equal.

Thus,$∠GAB$= $∠GDE$.

Then,$∠GAB$= $∠GED$.

The congruent angles to role="math" $∠GAB$are $∠GED,∠GDE,∠GBA$.