Let $ABCD$ be a parallelogram. Let $E$ and $F$ be the midpoints of $AB$ and $BC$ respectively. The lines $EC$ and $FD$ intersect in $P$ and form four triangles $APB$, $BPC$, $CPD$, and $DPA$. If the area of the parallelogram is $100$ sq. units, what is the maximum area in sq. units of a triangle among these four triangles?

Step 1 (Setup and Coordinates): Let the parallelogram $ABCD$ be represented in a coordinate system. Without loss of generality, we can scale and shear the coordinates so that:

• $A=(0,0)$

• $B=(1,0)$

• $D=(u,v)$ where the area of the parallelogram is given by $v=100$ sq. units.

• $C=(1+u,v)$

Step 2 (Midpoints and Equations of Lines):

• $E$ is the midpoint of $AB$, so $E=(0.5,0)$.

• $F$ is the midpoint of $BC$, so $F=(1+0.5u,0.5v)$.

Now we write the equations of the lines $EC$ and $FD$:

1. The line $EC$ passes through $E(0.5,0)$ and $C(1+u,v)$. The slope is $\frac{2v}{2u+1}$. The equation of line $EC$ is:

2. The line $FD$ passes through $D(u,v)$ and $F(1+0.5u,0.5v)$. The slope is $\frac{v}{u-2}$. The equation of line $FD$ is:

Step 3 (Find Intersection Point P): Equating the two line equations to find the intersection point $P(x_P,y_P)$:

Dividing by $v$ and solving for $x_P$:

Substituting $x_P$ back to find the height $y_P$:

Step 4 (Compute Areas of the Four Triangles): The height of $P$ relative to the base $AB$ is $y_P=0.6v$. The height relative to $CD$ is $v-0.6v=0.4v$.

1. Area of $△APB$: The base is $AB=1$. The height is $y_P=0.6v$.

2. Area of $△CPD$: The base is $CD=1$. The height is $0.4v$.

3. Area of $△BPC$: Using the triangle vertex determinant formula for $B(1,0)$, $C(1+u,v)$, and $P(x_P,y_P)$:

4. Area of $△DPA$: Since the total area of the parallelogram is $100$:

Step 5 (Conclusion): The areas of the four triangles are $10,20,30,40$. The maximum area is exactly $40$ sq. units.