Cannot prove a geometry area ratio between a triangle and a parallelogram ABCD is a parallelogram.
Prove the following:

1. $$\frac{BF}{FA} = \frac{AD}{AE}$$

2. $$\frac{S_{ADF}}{S_{AEF}} = \frac{AD}{AE}$$

3. $$S_{EBF} = S_{ADF}$$

4. $$S_{BCE} = \frac{1}{2}S_{ABCD}$$

I solved the first 3, but could not solve the 4th:

1.

$$\text{Thale's theorm:}$$ $$\frac{AE}{CB} = \frac{AF}{FB} = \frac{EF}{FC}$$ $$\frac{AE}{AD} = \frac{EF}{FC}$$ $$\downarrow$$ $$\frac{AE}{CB} = \frac{AF}{FB} = \frac{EF}{FC} = \frac{AE}{AD}$$ $$\frac{AF}{FB} = \frac{AE}{AD}$$ $$\downarrow$$ $$\boxed{\frac{FB}{AF} = \frac{AD}{AE}}$$

2.

$$\text{Let P be a point on ED such that FP will be perpendicular to ED.}$$ $$\div\begin{cases} S_{AEF} = \frac{AE\cdot FP}{2} \\ S_{ADF} = \frac{AD\cdot FP}{2}\end{cases}$$ $$\frac{S_{AEF}}{S_{ADF}} =\frac{AE}{AD}$$ $$\downarrow$$ $$\boxed{\frac{S_{ADF}}{S_{AEF}} =\frac{AD}{AE}}$$

3. $$\text{Let G be a point on AB such that EG will be perpendicular to AB. Then:}$$ $$\frac{S_{AEF}}{S_{EFB}} = \frac{\frac{AF\cdot EG}{2}}{\frac{FB\cdot EG}{2}} = \frac{AF}{FB} = \frac{AE}{AD} = \frac{S_{AEF}}{S_{ADF}}$$ $$\frac{S_{AEF}}{S_{EFB}} = \frac{S_{AEF}}{S_{ADF}}$$ $$\downarrow$$ $$\frac{S_{EFB}}{S_{AEF}} = \frac{S_{ADF}}{S_{AEF}}$$ $$\boxed{S_{EFB} = S_{ADF}}$$

1. I have absolutely no clue. I see no way of creating a relation between the areas of the triangle and the parallelogram. I thought of trying to somehow prove that the area of BCE is identical to that of BCD or BAD, but couldn't find a way to connect those either, as they don't have a shared perpendicular.

AD is parallel to BC. Therefore, perpendiculars from A and E to BC will have the same length, call it $$h$$. Thus, $$S_{BCE} = S_{BCA} = \frac{1}{2}h BC$$. But $$S_{BCA} = \frac{1}{2} S_{ABCD}$$ as it is exactly half of the parallelogram, and the statement follows.