4
$\begingroup$

enter image description here

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.
$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Sometimes I resent myself for missing something so seemingly obvious... $\endgroup$ – daedsidog Jan 3 at 23:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.