Extension and new area

Connect $B'C$ .

We have $[BB'C]=[ABC]$ since $AB=BB'$ .

We also have $[B'CC']=[BB'C]$ since $BC=CC'$ .

Similarly, we have $[ABC]=[BB'C]=[CC'A]=[AA'B]=[B'CC']=[C'AA']=[A'BB]$

However, $[A'B'C']=[ABC]+[BB'C]+[CC'A]+[AA'B]+[B'CC']+[C'AA']+[A'BB]$ , hence the answer is 7.

Let $T$ be the area of $\triangle ABC$ .

If $AA' = aCA$ , then $\triangle BAA'$ has the same height as $\triangle ABC$ but a base of $aCA$ instead of $CA$ , so $\triangle BAA'$ has an area of $aT$ .

If $BB' = bAB$ , then $\triangle CBB'$ has the same height as $\triangle ABC$ but a base of $bAB$ instead of $AB$ , so $\triangle CBB'$ has an area of $bT$ .

If $CC' = cBC$ , then $\triangle ACC'$ has the same height as $\triangle ABC$ but a base of $cBC$ instead of $AB$ , so $\triangle ACC'$ has an area of $cT$ .

Also if $AA' = aCA$ , then $\triangle C'AA'$ has the same height as $\triangle ACC'$ but a base of $aCA$ instead of $CA$ , so $\triangle C'AA'$ has an area of $acT$ .

Also if $BB' = bAB$ , then $\triangle A'BB'$ has the same height as $\triangle BAA'$ but a base of $bAB$ instead of $AB$ , so $\triangle A'BB'$ has an area of $abT$ .

Also if $CC' = cBC$ , then $\triangle B'CC'$ has the same height as $\triangle CBB'$ but a base of $cBC$ instead of $BC$ , so $\triangle B'CC'$ has an area of $bcT$ .

Therefore, the ratio of areas of $\triangle A'B'C'$ to $\triangle ABC$ is $\frac{T + aT + bT + cT + abT + acT + bcT}{T} = 1 + a + b + c + ab + ac + bc$ .

In this case, $a = b = c = 1$ , so the ratio of areas is $1 + 1 + 1 + 1 + 1 \cdot 1 + 1 \cdot 1 + 1 \cdot 1 = \boxed{7}$ .