http://209.85.48.8/9963/162/upload/p2206487.gif

The area must be equal.

How do I do this?

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- Jul 10th 2010, 10:00 PMCthul[Trigonometric Ratios] Dividing a triangle with equal area
http://209.85.48.8/9963/162/upload/p2206487.gif

The area must be equal.

How do I do this? - Jul 10th 2010, 10:18 PMAlso sprach Zarathustra
Use median from point C to AB (why?)

- Jul 10th 2010, 10:20 PMCthul
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 PMAlso sprach Zarathustra
- Jul 10th 2010, 10:43 PMCthul
Thanks. I didn't know finding the median will make the area still equivalent.

- Jul 10th 2010, 10:49 PMundefined
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 AMSoroban
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.