Prove that the area of the triangle formed by joining the midpoints of the sides of a triangle is one-fourth the area of the original triangle.

(N/A) Let the vertices of the triangle be $A(x_1, y_1)$,$B(x_2, y_2)$,and $C(x_3, y_3)$.
The area of $\triangle ABC$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.
Let $D, E,$ and $F$ be the midpoints of sides $BC, AC,$ and $AB$ respectively.
The coordinates of the midpoints are $D = (\frac{x_2+x_3}{2}, \frac{y_2+y_3}{2})$,$E = (\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2})$,and $F = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$.
Using the area formula for $\triangle DEF$,we calculate the area as $\frac{1}{2} |x_D(y_E - y_F) + x_E(y_F - y_D) + x_F(y_D - y_E)|$.
Substituting the coordinates and simplifying,we get the area of $\triangle DEF = \frac{1}{4} \times \text{Area of } \triangle ABC$.
Alternatively,by the Midpoint Theorem,$\triangle AFE \sim \triangle ABC$ with ratio $1:2$,so $\text{Area}(\triangle AFE) = \frac{1}{4} \text{Area}(\triangle ABC)$. Similarly for other small triangles,the central triangle $\triangle DEF$ also has an area equal to $\frac{1}{4} \text{Area}(\triangle ABC)$.