Hello, dudeosu!

It's easier than you think . . .

$\displaystyle \text{Suppose }ABC\text{ is a right triangle with }\angle C = 90^o.$

$\displaystyle \text{Let }M\text{ be the midpoint of the segment }AB.$

$\displaystyle \text{Prove that the area of }\Delta AMC\text{ is equal to the area of }\Delta MCB.$

Code:

C
*
*:\ *
* : \ *
* :h \ *
* : \ *
* : \ *
A * - - + - - * - - - - - * B
M

Since $\displaystyle \,M$ is the midpoint of $\displaystyle AB\!:\; AM = MB$

. . The two triangles have equal bases.

The two triangles have the same height, $\displaystyle \,h.$

Therefore, their areas are equal.

Note: This is true for *any* triangle.

. . . . .A median divides a triangle into two equal areas.