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

          oldest

          votes


















          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





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069699%2fwhere-does-this-property-involving-quadrilaterals-come-from%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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


















                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3069699%2fwhere-does-this-property-involving-quadrilaterals-come-from%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    "Incorrect syntax near the keyword 'ON'. (on update cascade, on delete cascade,)

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 total noob sorry any help greatly appreciated. I keep getting this "Incorrect syntax near the keyword 'ON'." heres my codes code one CREATE TABLE supplier ( supplierID INT PRIMARY KEY, ISBN INT FOREIGN KEY REFERENCES book(ISBN), price DECIMAL (5,2), ON UPDATE CASCADE, ON DELETE CASCADE, ); code two CREATE TABLE orders ( orderID INT PRIMARY KEY, customerID INT FOREIGN KEY REFERENCES customer(customerID), ISBN INT FOREIGN KEY REFERENCES book(ISBN), orderDate date, ON UPDATE CASCADE, ON DELETE CASCADE, ); Ive run out of stuff to try other than dumbing it down to uselessness.
                    Read more

                    Alcedinidae

                    Image
                    Alcedinidae Martin-pêcheur à dos bleu ( Alcedo azurea ) Classification (COI) Règne Animalia Embranchement Chordata Sous-embr. Vertebrata Classe Aves Ordre Coraciiformes Famille Alcedinidae Rafinesque, 1815 Les Alcédinidés ou Alcedinidae forment une famille d'oiseaux plus connus sous les noms de martins-pêcheurs et martins-chasseurs. Sommaire 1 Description 2 Répartition et habitat 3 Systématique 3.1 Liste alphabétique des genres 3.2 Liste des espèces 4 Chez les Anciens : l'alcyon 4.1 Étymologie (Littré) 4.2 Oiseau « fabuleux » ? 5 Notes 6 Articles connexes 7 Liens externes 7.1 Bibliographie Description | Ce sont des oiseaux compacts de taille petite à moyenne (10 à 46 cm), au bec droit et long, en forme de poignard. Leurs pattes sont courtes, et ils portent un plumage aux couleurs vives. Répartition et habitat | Cosmopolites, ils fréquentent surtout
                    Read more

                    RAC Tourist Trophy

                    Image
                    Pour les articles homonymes, voir Tourist Trophy. Le RAC Tourist Trophy est une course automobile organisée par le Royal Automobile Club d'Angleterre. Créée en 1905, cette course a fait partie des épreuves comptant pour le championnat du monde des voitures de sport entre 1953 et 1955, en 1958 et 1959, et entre 1962 et 1965. Après une interruption de 17 ans (l'épreuve s'arrête après 1988), le RAC Tourist Trophy reprend en 2005 sous l'égide de la FIA GT. Palmarès | John Napier sur Arrol-Johnston, vainqueur du premier RAC Tourist Trophy, en 1905. L'honorable C.S. Rolls, vainqueur du RAC Tourist Trophy 1906. Ernest Curtis, sur Rover 20 hp, vainqueur du RAC Tourist Trophy 1907. William Watson, vainqueur du RAC Tourist Trophy 1908. Kenelm Lee Guinness, vainqueur du RAC Tourist Trophy 1914 sur Sunbeam. Départ du RAC Tourist Trophy 1928. Louis Gérard, vainqueur du RAC Tourist Trophy 1938 sur Delage. Année Circuit Vainqueur
                    Read more