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Let $ABC$ be a triangle with sides $a,b,c$ and $A_1B_1C_1$ be another triangle with sides $a+\frac{b}2$, $b+\frac{c}2$, $c+\frac{a}2$. Prove that:
$$\frac94[ABC]=[A_1B_1C_1]$$
I tried using Heron's formula, but it is not getting me to the answer. Can anyone help? :)