Cannot prove a geometry area ratio between a triangle and a parallelogram












4














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.



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



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




  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.










share|cite|improve this question





























    4














    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.



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



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




    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.










    share|cite|improve this question



























      4












      4








      4







      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.



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



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




      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.










      share|cite|improve this question















      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.



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



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




      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.







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      edited Jan 3 at 23:15

























      asked Jan 3 at 23:03









      daedsidog

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          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.






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            Sometimes I resent myself for missing something so seemingly obvious...
            – daedsidog
            Jan 3 at 23:19











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          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.






          share|cite|improve this answer

















          • 1




            Sometimes I resent myself for missing something so seemingly obvious...
            – daedsidog
            Jan 3 at 23:19
















          3














          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.






          share|cite|improve this answer

















          • 1




            Sometimes I resent myself for missing something so seemingly obvious...
            – daedsidog
            Jan 3 at 23:19














          3












          3








          3






          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 23:16









          Todor Markov

          1,51839




          1,51839








          • 1




            Sometimes I resent myself for missing something so seemingly obvious...
            – daedsidog
            Jan 3 at 23:19














          • 1




            Sometimes I resent myself for missing something so seemingly obvious...
            – daedsidog
            Jan 3 at 23:19








          1




          1




          Sometimes I resent myself for missing something so seemingly obvious...
          – daedsidog
          Jan 3 at 23:19




          Sometimes I resent myself for missing something so seemingly obvious...
          – daedsidog
          Jan 3 at 23:19


















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