# Thread: [Trigonometric Ratios] Dividing a triangle with equal area

1. ## [Trigonometric Ratios] Dividing a triangle with equal area

The area must be equal.
How do I do this?

2. Use median from point C to AB (why?)

3. Oops, I forgot to mention the triangle has to have equal area.
Will the area be the same if I just divide it like that?

4. Originally Posted by Cthul
Oops, I forgot to mention the triangle has to have equal area.
I have considered that already...

5. Thanks. I didn't know finding the median will make the area still equivalent.

6. Originally Posted by Cthul
Thanks. I didn't know finding the median will make the area still equivalent.
There's a proof here. (The proof is for a stronger result.) This is related to the centroid being centre of mass for a thin triangular slab with uniform density.

7. Hello, Cthul!

"Also Sprach" is absolutely correct.

We have $\displaystyle \Delta ABC$ with median $\displaystyle AM$ drawn to side $\displaystyle BC.$
. . Hence: .$\displaystyle BM \,=\,MC.$

Draw altitude $\displaystyle h$ from vertex $\displaystyle A$ to side $\displaystyle BC.$

Code:
                A
o
*:* *
* : *   *
*  :  *     *
*   :   *       *
*    :h   *         *
*     :     *           *
*      :      *             *
*       :       *               *
*        :        *                 *
B o * * * * + * * * * o * * * * * * * * * o C
M

Triangles $\displaystyle ABM$ and $\displaystyle AMC$ have equal bases and the same altitude.
. . Hence, they have equal areas.

A median always divides a triangle into two equal areas.