Area of triangle within a triangle

Given a triangle ABC. On the sides: Let R lie one third of the way from A to B. Let S lie one third of the way from B to C. Let T lie one third of the way from C to A. Now connect points RST to make a triangle RST within triangle ABC. Prove that the area of triangle RST is equal to one third the area of triangle ABC.

Any help please?? I've tried law of sines and such to no avail.

Re: Area of triangle within a triangle

The area *A* of triangle ABC is:

$\displaystyle A=\frac{1}{6}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)$

The area $\displaystyle A_O$ within ABC but outside of RST is:

$\displaystyle A_O=\frac{1}{9}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)$

Hence the area within RST is:

$\displaystyle A-A_O=\frac{1}{18}\left(bc\sin(A)+ac\sin(B)+ab\sin(C ) \right)=\frac{1}{3}A$

Re: Area of triangle within a triangle

Where did the 1/6 and 1/9 come from?

Re: Area of triangle within a triangle

I used the formula for the area of a triangle:

$\displaystyle A=\frac{1}{2}ab\sin(C)$

Now, for triangle ABC, I included all three angles so that each term is 1/3 of the entire area, or equivalently the area of ABC can be found from:

$\displaystyle 3A=\frac{1}{2}bc\sin(A)+\frac{1}{2}ac\sin(B)+\frac {1}{2}ab\sin(C)$ and so:

$\displaystyle A=\frac{1}{6}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)$

For the 3 triangles outside of RST, I used the same formula:

$\displaystyle A_O=\frac{1}{2}\cdot\frac{2}{3}b\cdot\frac{1}{3}c \sin(A)+\frac{1}{2}\cdot\frac{1}{3}a\cdot\frac{2}{ 3}c\sin(B)+\frac{1}{2}\cdot\frac{2}{3}a\cdot\frac{ 1}{3}b\sin(C)$

$\displaystyle A_O=\frac{1}{9}\left(bc\sin(A)+ac\sin(B)+ab\sin(C) \right)$