Medians, medians and more medians

$A_i$ and $A_{i-1}$ are related by the formula $A_{i}=\displaystyle\frac{3A_{i-1}}{4}$

Thus $\displaystyle\sum_{i=0}^{\infty}A_i=A_0\displaystyle\sum_{n=0}^{\infty}\displaystyle(\frac{3}{4})^{n}=4A_0$

Proof that $A_{i}=\displaystyle\frac{3A_{i-1}}{4}$

$\square BGCG'$ is a parallelogram with sides $2\beta, 2\gamma, 2\beta, 2\gamma$ , Thus area of $\triangle GG'C$ is equal to the area of $\triangle BGC$

Now let the area of triangle with sides $\alpha, \beta, \gamma$ be $\Delta$ .

Thus area of $\triangle GG'C$ which has sides $2\alpha, 2\beta, 2\gamma$ is $4\Delta$ , Thus area of $\triangle BGC$ is $4\Delta$ $(If \ sides \ are \ scaled \ by \ a \ factor \ of \ k \ areas \ are \ scaled \ by \ a \ factor \ of \ k^{2})$

By similar reasoning, we can say areas of $\triangle BGC \ \triangle CGA \ \& \ \triangle AGB$ are all $4\Delta$ .

Thus area of $\triangle ABC$ is $12\Delta$

Also, the medians have lengths $3\alpha, 3\beta, 3\gamma$ , Thus the triangle formed by the medians will have an area of $9\Delta$

Thus the area of triangle $\Delta_A=12\Delta$ .

And the area of the triangle formed by medians $\Delta_M=9\Delta$ .

Thus we get the relation $\Delta_M=\displaystyle\frac{3\Delta_A}{4}$ .