Where does this property involving quadrilaterals come from?












4












$begingroup$


enter image description here




$ABCD$ is a square. $|AF|=6$, $|FK|=2$, and $DE parallel AB$. What is $|EK|=?$




My geometry book has a property for this:



$$|AF|^2=|FK|cdot|FE|$$



Can you show me where does this property come from in simple terms?






geometry quadrilateral







share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    enter image description here




    $ABCD$ is a square. $|AF|=6$, $|FK|=2$, and $DE parallel AB$. What is $|EK|=?$




    My geometry book has a property for this:



    $$|AF|^2=|FK|cdot|FE|$$



    Can you show me where does this property come from in simple terms?






    geometry quadrilateral







    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      enter image description here




      $ABCD$ is a square. $|AF|=6$, $|FK|=2$, and $DE parallel AB$. What is $|EK|=?$




      My geometry book has a property for this:



      $$|AF|^2=|FK|cdot|FE|$$



      Can you show me where does this property come from in simple terms?






      geometry quadrilateral







      share|cite|improve this question











      $endgroup$




      enter image description here




      $ABCD$ is a square. $|AF|=6$, $|FK|=2$, and $DE parallel AB$. What is $|EK|=?$




      My geometry book has a property for this:



      $$|AF|^2=|FK|cdot|FE|$$



      Can you show me where does this property come from in simple terms?






      geometry quadrilateral




      geometry quadrilateral






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      steven gregory

      17.8k32257




      17.8k32257










      asked 2 days ago









      Eldar RahimliEldar Rahimli

      1088




      1088






















          2 Answers
          2






          active

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          7












          $begingroup$

          $triangle AFD$ is similar to$triangle KFB$ so $|AF|/|KF|=|FD|/|FB|$.



          $triangle AFB$ is similar to$triangle EFD$ so $|EF|/|AF|=|FD|/|FB|$.



          So $|EF|/|AF|=|AF|/|KF|$ proving the claim.



          You don't need a square. All you need is a parallelogram, the parallel sides enable the angle congruence that make the relevant triangles similar.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            Prove that $FC$ is a tangent line to the circumcircle of $Delta KCE$, for which prove that
            $$measuredangle CEK=measuredangle FAB=measuredangle FCK.$$
            After this use $AF=CF$ and $$FC^2=FKcdot FC.$$






            share|cite|improve this answer











            $endgroup$









            • 1




              $begingroup$
              $CF$ is tangent, maybe?
              $endgroup$
              – Oscar Lanzi
              2 days ago






            • 1




              $begingroup$
              @Oscar Lanz It was typo. Thank you!
              $endgroup$
              – Michael Rozenberg
              2 days ago











            Your Answer





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            2 Answers
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            2 Answers
            2






            active

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            active

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            active

            oldest

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            7












            $begingroup$

            $triangle AFD$ is similar to$triangle KFB$ so $|AF|/|KF|=|FD|/|FB|$.



            $triangle AFB$ is similar to$triangle EFD$ so $|EF|/|AF|=|FD|/|FB|$.



            So $|EF|/|AF|=|AF|/|KF|$ proving the claim.



            You don't need a square. All you need is a parallelogram, the parallel sides enable the angle congruence that make the relevant triangles similar.






            share|cite|improve this answer









            $endgroup$


















              7












              $begingroup$

              $triangle AFD$ is similar to$triangle KFB$ so $|AF|/|KF|=|FD|/|FB|$.



              $triangle AFB$ is similar to$triangle EFD$ so $|EF|/|AF|=|FD|/|FB|$.



              So $|EF|/|AF|=|AF|/|KF|$ proving the claim.



              You don't need a square. All you need is a parallelogram, the parallel sides enable the angle congruence that make the relevant triangles similar.






              share|cite|improve this answer









              $endgroup$
















                7












                7








                7





                $begingroup$

                $triangle AFD$ is similar to$triangle KFB$ so $|AF|/|KF|=|FD|/|FB|$.



                $triangle AFB$ is similar to$triangle EFD$ so $|EF|/|AF|=|FD|/|FB|$.



                So $|EF|/|AF|=|AF|/|KF|$ proving the claim.



                You don't need a square. All you need is a parallelogram, the parallel sides enable the angle congruence that make the relevant triangles similar.






                share|cite|improve this answer









                $endgroup$



                $triangle AFD$ is similar to$triangle KFB$ so $|AF|/|KF|=|FD|/|FB|$.



                $triangle AFB$ is similar to$triangle EFD$ so $|EF|/|AF|=|FD|/|FB|$.



                So $|EF|/|AF|=|AF|/|KF|$ proving the claim.



                You don't need a square. All you need is a parallelogram, the parallel sides enable the angle congruence that make the relevant triangles similar.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Oscar LanziOscar Lanzi

                12.3k12036




                12.3k12036























                    2












                    $begingroup$

                    Prove that $FC$ is a tangent line to the circumcircle of $Delta KCE$, for which prove that
                    $$measuredangle CEK=measuredangle FAB=measuredangle FCK.$$
                    After this use $AF=CF$ and $$FC^2=FKcdot FC.$$






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      $CF$ is tangent, maybe?
                      $endgroup$
                      – Oscar Lanzi
                      2 days ago






                    • 1




                      $begingroup$
                      @Oscar Lanz It was typo. Thank you!
                      $endgroup$
                      – Michael Rozenberg
                      2 days ago
















                    2












                    $begingroup$

                    Prove that $FC$ is a tangent line to the circumcircle of $Delta KCE$, for which prove that
                    $$measuredangle CEK=measuredangle FAB=measuredangle FCK.$$
                    After this use $AF=CF$ and $$FC^2=FKcdot FC.$$






                    share|cite|improve this answer











                    $endgroup$









                    • 1




                      $begingroup$
                      $CF$ is tangent, maybe?
                      $endgroup$
                      – Oscar Lanzi
                      2 days ago






                    • 1




                      $begingroup$
                      @Oscar Lanz It was typo. Thank you!
                      $endgroup$
                      – Michael Rozenberg
                      2 days ago














                    2












                    2








                    2





                    $begingroup$

                    Prove that $FC$ is a tangent line to the circumcircle of $Delta KCE$, for which prove that
                    $$measuredangle CEK=measuredangle FAB=measuredangle FCK.$$
                    After this use $AF=CF$ and $$FC^2=FKcdot FC.$$






                    share|cite|improve this answer











                    $endgroup$



                    Prove that $FC$ is a tangent line to the circumcircle of $Delta KCE$, for which prove that
                    $$measuredangle CEK=measuredangle FAB=measuredangle FCK.$$
                    After this use $AF=CF$ and $$FC^2=FKcdot FC.$$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 2 days ago

























                    answered 2 days ago









                    Michael RozenbergMichael Rozenberg

                    97.9k1590188




                    97.9k1590188








                    • 1




                      $begingroup$
                      $CF$ is tangent, maybe?
                      $endgroup$
                      – Oscar Lanzi
                      2 days ago






                    • 1




                      $begingroup$
                      @Oscar Lanz It was typo. Thank you!
                      $endgroup$
                      – Michael Rozenberg
                      2 days ago














                    • 1




                      $begingroup$
                      $CF$ is tangent, maybe?
                      $endgroup$
                      – Oscar Lanzi
                      2 days ago






                    • 1




                      $begingroup$
                      @Oscar Lanz It was typo. Thank you!
                      $endgroup$
                      – Michael Rozenberg
                      2 days ago








                    1




                    1




                    $begingroup$
                    $CF$ is tangent, maybe?
                    $endgroup$
                    – Oscar Lanzi
                    2 days ago




                    $begingroup$
                    $CF$ is tangent, maybe?
                    $endgroup$
                    – Oscar Lanzi
                    2 days ago




                    1




                    1




                    $begingroup$
                    @Oscar Lanz It was typo. Thank you!
                    $endgroup$
                    – Michael Rozenberg
                    2 days ago




                    $begingroup$
                    @Oscar Lanz It was typo. Thank you!
                    $endgroup$
                    – Michael Rozenberg
                    2 days ago


















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