In the square $ABCD$, prove that $BF+DE=AE$.












5












$begingroup$


Consider the Square $ABCD$. The point $E$ is on the side $CD$. If $F$ is on the side $BC$ such that $AF$ is the bisector of the angle $BAE$. Prove that
$$BF+DE=AE.$$
By Pythagorean theorem we have $(AD)^2+(DE)^2=(AE)^2$ then $DE=sqrt{(AE)^2-(AD)^2}$ also, $DE=AEsin(EAD)$ and $BF=AFsin(BAF)$ on the other hand we have $BAE+EAD=90^0$






calculus geometry area







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your question?
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:18










  • $begingroup$
    @Théophile, prove the mentioned identity is my question.
    $endgroup$
    – soodehMehboodi
    Nov 23 '18 at 19:20










  • $begingroup$
    Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:26










  • $begingroup$
    Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
    $endgroup$
    – soodehMehboodi
    Nov 23 '18 at 19:31










  • $begingroup$
    Good. Please add an explanation of your work so far to make it easier to answer your question.
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:32
















5












$begingroup$


Consider the Square $ABCD$. The point $E$ is on the side $CD$. If $F$ is on the side $BC$ such that $AF$ is the bisector of the angle $BAE$. Prove that
$$BF+DE=AE.$$
By Pythagorean theorem we have $(AD)^2+(DE)^2=(AE)^2$ then $DE=sqrt{(AE)^2-(AD)^2}$ also, $DE=AEsin(EAD)$ and $BF=AFsin(BAF)$ on the other hand we have $BAE+EAD=90^0$






calculus geometry area







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is your question?
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:18










  • $begingroup$
    @Théophile, prove the mentioned identity is my question.
    $endgroup$
    – soodehMehboodi
    Nov 23 '18 at 19:20










  • $begingroup$
    Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:26










  • $begingroup$
    Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
    $endgroup$
    – soodehMehboodi
    Nov 23 '18 at 19:31










  • $begingroup$
    Good. Please add an explanation of your work so far to make it easier to answer your question.
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:32














5












5








5


3



$begingroup$


Consider the Square $ABCD$. The point $E$ is on the side $CD$. If $F$ is on the side $BC$ such that $AF$ is the bisector of the angle $BAE$. Prove that
$$BF+DE=AE.$$
By Pythagorean theorem we have $(AD)^2+(DE)^2=(AE)^2$ then $DE=sqrt{(AE)^2-(AD)^2}$ also, $DE=AEsin(EAD)$ and $BF=AFsin(BAF)$ on the other hand we have $BAE+EAD=90^0$






calculus geometry area







share|cite|improve this question











$endgroup$




Consider the Square $ABCD$. The point $E$ is on the side $CD$. If $F$ is on the side $BC$ such that $AF$ is the bisector of the angle $BAE$. Prove that
$$BF+DE=AE.$$
By Pythagorean theorem we have $(AD)^2+(DE)^2=(AE)^2$ then $DE=sqrt{(AE)^2-(AD)^2}$ also, $DE=AEsin(EAD)$ and $BF=AFsin(BAF)$ on the other hand we have $BAE+EAD=90^0$






calculus geometry area




calculus geometry area






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 '18 at 19:49







soodehMehboodi

















asked Nov 23 '18 at 19:16









soodehMehboodisoodehMehboodi

55428




55428












  • $begingroup$
    What is your question?
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:18










  • $begingroup$
    @Théophile, prove the mentioned identity is my question.
    $endgroup$
    – soodehMehboodi
    Nov 23 '18 at 19:20










  • $begingroup$
    Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:26










  • $begingroup$
    Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
    $endgroup$
    – soodehMehboodi
    Nov 23 '18 at 19:31










  • $begingroup$
    Good. Please add an explanation of your work so far to make it easier to answer your question.
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:32


















  • $begingroup$
    What is your question?
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:18










  • $begingroup$
    @Théophile, prove the mentioned identity is my question.
    $endgroup$
    – soodehMehboodi
    Nov 23 '18 at 19:20










  • $begingroup$
    Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:26










  • $begingroup$
    Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
    $endgroup$
    – soodehMehboodi
    Nov 23 '18 at 19:31










  • $begingroup$
    Good. Please add an explanation of your work so far to make it easier to answer your question.
    $endgroup$
    – Théophile
    Nov 23 '18 at 19:32
















$begingroup$
What is your question?
$endgroup$
– Théophile
Nov 23 '18 at 19:18




$begingroup$
What is your question?
$endgroup$
– Théophile
Nov 23 '18 at 19:18












$begingroup$
@Théophile, prove the mentioned identity is my question.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:20




$begingroup$
@Théophile, prove the mentioned identity is my question.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:20












$begingroup$
Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
$endgroup$
– Théophile
Nov 23 '18 at 19:26




$begingroup$
Well, it's not written as a question, it's more of an order. Have you done any work on this yourself?
$endgroup$
– Théophile
Nov 23 '18 at 19:26












$begingroup$
Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:31




$begingroup$
Yes I drew the shape and use some properties of bisectors also Applying Pythagorean theorem, but I did not reach to aim.
$endgroup$
– soodehMehboodi
Nov 23 '18 at 19:31












$begingroup$
Good. Please add an explanation of your work so far to make it easier to answer your question.
$endgroup$
– Théophile
Nov 23 '18 at 19:32




$begingroup$
Good. Please add an explanation of your work so far to make it easier to answer your question.
$endgroup$
– Théophile
Nov 23 '18 at 19:32










2 Answers
2






active

oldest

votes


















4












$begingroup$

Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)



enter image description here






share|cite|improve this answer











$endgroup$





















    3












    $begingroup$

    Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
    $$
    \BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
    \AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
    \BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
    \cos2alpha=1-2sin^2alpha
    $$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

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      4












      $begingroup$

      Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)



      enter image description here






      share|cite|improve this answer











      $endgroup$


















        4












        $begingroup$

        Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)



        enter image description here






        share|cite|improve this answer











        $endgroup$
















          4












          4








          4





          $begingroup$

          Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)



          enter image description here






          share|cite|improve this answer











          $endgroup$



          Hint: Rotate the $F$ around $A$ for $90^{circ}$ (you get new point name it $H$) and $B$ goes to $D$. Note that $E,D,H$ are collinear. Now use congruence theorems. (Prove that $HE =EA$.)



          enter image description here







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 23 '18 at 20:02

























          answered Nov 23 '18 at 19:53









          greedoidgreedoid

          38.6k114797




          38.6k114797























              3












              $begingroup$

              Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
              $$
              \BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
              \AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
              \BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
              \cos2alpha=1-2sin^2alpha
              $$






              share|cite|improve this answer









              $endgroup$


















                3












                $begingroup$

                Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
                $$
                \BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
                \AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
                \BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
                \cos2alpha=1-2sin^2alpha
                $$






                share|cite|improve this answer









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
                  $$
                  \BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
                  \AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
                  \BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
                  \cos2alpha=1-2sin^2alpha
                  $$






                  share|cite|improve this answer









                  $endgroup$



                  Let $AB=x$ and angle $<BAF=<FAE=alpha=> DE=xtan(frac{pi}{2}-2alpha),; BF=xtanalpha=>$
                  $$
                  \BF+DE=x(tan(frac{pi}{2}-2alpha)+tanalpha)=x(cot2alpha+tanalpha)
                  \AE=frac{x}{cos(frac{pi}{2}-2alpha)}=frac{x}{sin2alpha}
                  \BF+DE=AE<=>cot2alpha+tanalpha=frac{1}{sin2alpha}<=>
                  \cos2alpha=1-2sin^2alpha
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 23 '18 at 19:53









                  Samvel SafaryanSamvel Safaryan

                  501111




                  501111






























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