# [Trigonometric Ratios] Dividing a triangle with equal area

• Jul 10th 2010, 10:00 PM
Cthul
[Trigonometric Ratios] Dividing a triangle with equal area
The area must be equal.
How do I do this?
• Jul 10th 2010, 10:18 PM
Also sprach Zarathustra
Use median from point C to AB (why?)
• Jul 10th 2010, 10:20 PM
Cthul
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?
• Jul 10th 2010, 10:21 PM
Also sprach Zarathustra
Quote:

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

• Jul 10th 2010, 10:43 PM
Cthul
Thanks. I didn't know finding the median will make the area still equivalent.
• Jul 10th 2010, 10:49 PM
undefined
Quote:

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.
• Jul 11th 2010, 04:54 AM
Soroban
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.