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Cannot prove a geometry area ratio between a triangle and a parallelogram
$begingroup$
ABCD is a parallelogram.
Prove the following:
$frac{BF}{FA} = frac{AD}{AE}$
$frac{S_{ADF}}{S_{AEF}} = frac{AD}{AE}$
$S_{EBF} = S_{ADF}$
$S_{BCE} = frac{1}{2}S_{ABCD}$
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.
$$text{Let P be a point on ED such that FP will be perpendicular to ED.}$$
$$divbegin{cases} S_{AEF} = frac{AEcdot FP}{2} \ S_{ADF} = frac{ADcdot FP}{2}end{cases}$$
$$frac{S_{AEF}}{S_{ADF}} =frac{AE}{AD}$$
$$downarrow$$
$$boxed{frac{S_{ADF}}{S_{AEF}} =frac{AD}{AE}}$$
3.
$$text{Let G be a point on AB such that EG will be perpendicular to AB. Then:}$$
$$frac{S_{AEF}}{S_{EFB}} = frac{frac{AFcdot EG}{2}}{frac{FBcdot EG}{2}} = frac{AF}{FB} = frac{AE}{AD} = frac{S_{AEF}}{S_{ADF}}$$
$$frac{S_{AEF}}{S_{EFB}} = frac{S_{AEF}}{S_{ADF}}$$
$$downarrow$$
$$frac{S_{EFB}}{S_{AEF}} = frac{S_{ADF}}{S_{AEF}}$$
$$boxed{S_{EFB} = S_{ADF}}$$
- 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.
geometry
asked Jan 3 at 23:03
daedsidogdaedsidog
27617
$endgroup$
add a comment |
$begingroup$
ABCD is a parallelogram.
Prove the following:
$frac{BF}{FA} = frac{AD}{AE}$
$frac{S_{ADF}}{S_{AEF}} = frac{AD}{AE}$
$S_{EBF} = S_{ADF}$
$S_{BCE} = frac{1}{2}S_{ABCD}$
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.
$$text{Let P be a point on ED such that FP will be perpendicular to ED.}$$
$$divbegin{cases} S_{AEF} = frac{AEcdot FP}{2} \ S_{ADF} = frac{ADcdot FP}{2}end{cases}$$
$$frac{S_{AEF}}{S_{ADF}} =frac{AE}{AD}$$
$$downarrow$$
$$boxed{frac{S_{ADF}}{S_{AEF}} =frac{AD}{AE}}$$
3.
$$text{Let G be a point on AB such that EG will be perpendicular to AB. Then:}$$
$$frac{S_{AEF}}{S_{EFB}} = frac{frac{AFcdot EG}{2}}{frac{FBcdot EG}{2}} = frac{AF}{FB} = frac{AE}{AD} = frac{S_{AEF}}{S_{ADF}}$$
$$frac{S_{AEF}}{S_{EFB}} = frac{S_{AEF}}{S_{ADF}}$$
$$downarrow$$
$$frac{S_{EFB}}{S_{AEF}} = frac{S_{ADF}}{S_{AEF}}$$
$$boxed{S_{EFB} = S_{ADF}}$$
- 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.
geometry
asked Jan 3 at 23:03
daedsidogdaedsidog
27617
$endgroup$
add a comment |
$begingroup$
ABCD is a parallelogram.
Prove the following:
$frac{BF}{FA} = frac{AD}{AE}$
$frac{S_{ADF}}{S_{AEF}} = frac{AD}{AE}$
$S_{EBF} = S_{ADF}$
$S_{BCE} = frac{1}{2}S_{ABCD}$
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.
$$text{Let P be a point on ED such that FP will be perpendicular to ED.}$$
$$divbegin{cases} S_{AEF} = frac{AEcdot FP}{2} \ S_{ADF} = frac{ADcdot FP}{2}end{cases}$$
$$frac{S_{AEF}}{S_{ADF}} =frac{AE}{AD}$$
$$downarrow$$
$$boxed{frac{S_{ADF}}{S_{AEF}} =frac{AD}{AE}}$$
3.
$$text{Let G be a point on AB such that EG will be perpendicular to AB. Then:}$$
$$frac{S_{AEF}}{S_{EFB}} = frac{frac{AFcdot EG}{2}}{frac{FBcdot EG}{2}} = frac{AF}{FB} = frac{AE}{AD} = frac{S_{AEF}}{S_{ADF}}$$
$$frac{S_{AEF}}{S_{EFB}} = frac{S_{AEF}}{S_{ADF}}$$
$$downarrow$$
$$frac{S_{EFB}}{S_{AEF}} = frac{S_{ADF}}{S_{AEF}}$$
$$boxed{S_{EFB} = S_{ADF}}$$
- 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.
geometry
asked Jan 3 at 23:03
daedsidogdaedsidog
27617
$endgroup$
ABCD is a parallelogram.
Prove the following:
$frac{BF}{FA} = frac{AD}{AE}$
$frac{S_{ADF}}{S_{AEF}} = frac{AD}{AE}$
$S_{EBF} = S_{ADF}$
$S_{BCE} = frac{1}{2}S_{ABCD}$
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.
$$text{Let P be a point on ED such that FP will be perpendicular to ED.}$$
$$divbegin{cases} S_{AEF} = frac{AEcdot FP}{2} \ S_{ADF} = frac{ADcdot FP}{2}end{cases}$$
$$frac{S_{AEF}}{S_{ADF}} =frac{AE}{AD}$$
$$downarrow$$
$$boxed{frac{S_{ADF}}{S_{AEF}} =frac{AD}{AE}}$$
3.
$$text{Let G be a point on AB such that EG will be perpendicular to AB. Then:}$$
$$frac{S_{AEF}}{S_{EFB}} = frac{frac{AFcdot EG}{2}}{frac{FBcdot EG}{2}} = frac{AF}{FB} = frac{AE}{AD} = frac{S_{AEF}}{S_{ADF}}$$
$$frac{S_{AEF}}{S_{EFB}} = frac{S_{AEF}}{S_{ADF}}$$
$$downarrow$$
$$frac{S_{EFB}}{S_{AEF}} = frac{S_{ADF}}{S_{AEF}}$$
$$boxed{S_{EFB} = S_{ADF}}$$
- 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.
geometry
geometry
asked Jan 3 at 23:03
daedsidogdaedsidog
27617
asked Jan 3 at 23:03
daedsidogdaedsidog
27617
edited Jan 3 at 23:15
daedsidog
asked Jan 3 at 23:03
daedsidogdaedsidog
27617
asked Jan 3 at 23:03
daedsidogdaedsidog
27617
asked Jan 3 at 23:03
daedsidogdaedsidog
27617
27617
add a comment |
add a comment |
1 Answer
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$begingroup$
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.
answered Jan 3 at 23:16
Todor MarkovTodor Markov
1,819410
$endgroup$
1
$begingroup$
Sometimes I resent myself for missing something so seemingly obvious...
$endgroup$
– daedsidog
Jan 3 at 23:19
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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.
answered Jan 3 at 23:16
Todor MarkovTodor Markov
1,819410
$endgroup$
1
$begingroup$
Sometimes I resent myself for missing something so seemingly obvious...
$endgroup$
– daedsidog
Jan 3 at 23:19
add a comment |
$begingroup$
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.
answered Jan 3 at 23:16
Todor MarkovTodor Markov
1,819410
$endgroup$
1
$begingroup$
Sometimes I resent myself for missing something so seemingly obvious...
$endgroup$
– daedsidog
Jan 3 at 23:19
add a comment |
$begingroup$
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.
answered Jan 3 at 23:16
Todor MarkovTodor Markov
1,819410
$endgroup$
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.
answered Jan 3 at 23:16
Todor MarkovTodor Markov
1,819410
answered Jan 3 at 23:16
Todor MarkovTodor Markov
1,819410
answered Jan 3 at 23:16
Todor MarkovTodor Markov
1,819410
answered Jan 3 at 23:16
Todor MarkovTodor Markov
1,819410
1,819410
1
$begingroup$
Sometimes I resent myself for missing something so seemingly obvious...
$endgroup$
– daedsidog
Jan 3 at 23:19
add a comment |
1
$begingroup$
Sometimes I resent myself for missing something so seemingly obvious...
$endgroup$
– daedsidog
Jan 3 at 23:19
1
1
$begingroup$
Sometimes I resent myself for missing something so seemingly obvious...
$endgroup$
– daedsidog
Jan 3 at 23:19
$begingroup$
Sometimes I resent myself for missing something so seemingly obvious...
$endgroup$
– daedsidog
Jan 3 at 23:19
add a comment |
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