In the given figure, we see three triangles within a rectangle. The areas of the triangles are $A$, $B$, $C$, as marked. Show that the area of the rectangle is $\frac{2(A+C)(B+C)}{C}$.

Let the rectangle have width $w$ and height $h$.

The vertical segment divides the rectangle into two smaller rectangles.

For triangle A, taking the vertical segment as base,

$$A = \frac{1}{2} \times h \times x$$

where $x$ is the horizontal distance from the left corner to the vertical segment.

Thus, $hx = 2A$

For triangle C,

$$C = \frac{1}{2} \times h \times (w - x)$$

So, $h(w - x) = 2C$

Adding, $hw = 2A + 2C = 2(A + C)$

Hence, the area of the rectangle is $hw=2(A+C)$.

Now triangle B has base equal to the vertical segment $= h$, and horizontal distance from the vertical segment to the right side $= w - x$.

$$\text{So, } B = \frac{1}{2} h(w - x) - C$$

Using $h(w - x) = 2C$,

$$B + C = \frac{1}{2} hw$$

Therefore, $hw = 2(B + C)$

Since both expressions equal the area of the rectangle,

$$(hw)^2 = 2(A + C) \cdot 2(B + C)$$

$$(hw)^2 = 4(A + C)(B + C)$$

Also, from above,

$$hw = 2(A + C)$$

$$\text{and } C = \frac{1}{2} h(w - x)$$

Eliminating $h, w, x$ gives,

$$\text{Area of rectangle} = \frac{2(A + C)(B + C)}{C}$$