Understanding Ceva's Theorem Announcing the arrival of Valued Associate #679: Cesar Manara ...

What causes the vertical darker bands in my photo?

Why is "Consequences inflicted." not a sentence?

Single word antonym of "flightless"

Apollo command module space walk?

Fundamental Solution of the Pell Equation

51k Euros annually for a family of 4 in Berlin: Is it enough?

Understanding Ceva's Theorem

The logistics of corpse disposal

2001: A Space Odyssey's use of the song "Daisy Bell" (Bicycle Built for Two); life imitates art or vice-versa?

Is it true that "carbohydrates are of no use for the basal metabolic need"?

How can I make names more distinctive without making them longer?

Output the ŋarâþ crîþ alphabet song without using (m)any letters

How do I keep my slimes from escaping their pens?

Is it ethical to give a final exam after the professor has quit before teaching the remaining chapters of the course?

Should I discuss the type of campaign with my players?

Why do people hide their license plates in the EU?

What exactly is a "Meth" in Altered Carbon?

At the end of Thor: Ragnarok why don't the Asgardians turn and head for the Bifrost as per their original plan?

What is the meaning of the new sigil in Game of Thrones Season 8 intro?

When were vectors invented?

Identify plant with long narrow paired leaves and reddish stems

Dating a Former Employee

What does this icon in iOS Stardew Valley mean?

Okay to merge included columns on otherwise identical indexes?



Understanding Ceva's Theorem



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Ratio of areas of similar triangles given SSHow to calculate Fermat point in a triangle most efficiently?Prove this is a rectangleThe formula for the area of two triangles determined by the diagonals of a trapezoidUSAMO 2005, Problem3 (Triangle Geometry)- Is my solution correct?Corresponding side in similar trianglesMake isosceles triangle with matchsticksCutting a Triangle Through Its CentroidSimilar triangles and cross section integrals.Postulate or Theorem: The areas of similar plane figures are proportional their linear dimensions squared?












5












$begingroup$


enter image description here



In Ceva's Theorem, I understand that $frac{A_{triangle PXB}}{A_{triangle PXC}}=frac{BX}{CX}=frac{A_{triangle BXA}}{A_{triangle CXA}}$.



I would like clarification in understanding the following step which states:



$frac{A_{triangle APB}}{A_{triangle APC}}=frac{A_{triangle AXB} - A_{triangle PXB}}{A_{triangle AXC}-A_{triangle PXC}}=frac{BX}{CX}$



How does the subtraction of the two areas make it so that the new triangles are still proportional to $frac{BX}{CX}$? (even though they do not share those sides!)






geometry proof-verification triangles







share|cite|improve this question











$endgroup$

















    5












    $begingroup$


    enter image description here



    In Ceva's Theorem, I understand that $frac{A_{triangle PXB}}{A_{triangle PXC}}=frac{BX}{CX}=frac{A_{triangle BXA}}{A_{triangle CXA}}$.



    I would like clarification in understanding the following step which states:



    $frac{A_{triangle APB}}{A_{triangle APC}}=frac{A_{triangle AXB} - A_{triangle PXB}}{A_{triangle AXC}-A_{triangle PXC}}=frac{BX}{CX}$



    How does the subtraction of the two areas make it so that the new triangles are still proportional to $frac{BX}{CX}$? (even though they do not share those sides!)






    geometry proof-verification triangles







    share|cite|improve this question











    $endgroup$















      5












      5








      5


      1



      $begingroup$


      enter image description here



      In Ceva's Theorem, I understand that $frac{A_{triangle PXB}}{A_{triangle PXC}}=frac{BX}{CX}=frac{A_{triangle BXA}}{A_{triangle CXA}}$.



      I would like clarification in understanding the following step which states:



      $frac{A_{triangle APB}}{A_{triangle APC}}=frac{A_{triangle AXB} - A_{triangle PXB}}{A_{triangle AXC}-A_{triangle PXC}}=frac{BX}{CX}$



      How does the subtraction of the two areas make it so that the new triangles are still proportional to $frac{BX}{CX}$? (even though they do not share those sides!)






      geometry proof-verification triangles







      share|cite|improve this question











      $endgroup$




      enter image description here



      In Ceva's Theorem, I understand that $frac{A_{triangle PXB}}{A_{triangle PXC}}=frac{BX}{CX}=frac{A_{triangle BXA}}{A_{triangle CXA}}$.



      I would like clarification in understanding the following step which states:



      $frac{A_{triangle APB}}{A_{triangle APC}}=frac{A_{triangle AXB} - A_{triangle PXB}}{A_{triangle AXC}-A_{triangle PXC}}=frac{BX}{CX}$



      How does the subtraction of the two areas make it so that the new triangles are still proportional to $frac{BX}{CX}$? (even though they do not share those sides!)






      geometry proof-verification triangles




      geometry proof-verification triangles






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 3 hours ago









      YuiTo Cheng

      2,48341037




      2,48341037










      asked 3 hours ago









      dragonkingdragonking

      384




      384






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          $A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$



          $A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$



          Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$






          share|cite|improve this answer









          $endgroup$














            Your Answer








            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%2f3190589%2funderstanding-cevas-theorem%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            $A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$



            $A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$



            Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              $A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$



              $A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$



              Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                $A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$



                $A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$



                Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$






                share|cite|improve this answer









                $endgroup$



                $A_{triangle AXB}: A_{triangle AXC}=BX:CXRightarrow A_{triangle AXB}=frac{BX}{CX}A_{triangle AXC}$



                $A_{triangle PXB}: A_{triangle PXC}=BX:CXRightarrow A_{triangle PXB}=frac{BX}{CX}A_{triangle PXC}$



                Hence $$frac{A_{triangle A X B} -A _{triangle P X B}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{frac{BX}{CX}A_{triangle AXC}-frac{BX}{CX}A_{triangle PXC}}{A _{triangle A X C}-A_{ triangle P X C}}=frac{BX}{CX}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 hours ago









                YuiTo ChengYuiTo Cheng

                2,48341037




                2,48341037






























                    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%2f3190589%2funderstanding-cevas-theorem%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

                    Can't compile dgruyter and caption packagesLaTeX templates/packages for writing a patent specificationLatex...

                    Image
                    Why does Arabsat 6A need a Falcon Heavy to launch Etiquette around loan refinance - decision is going to cost first broker a lot of money What is the PIE reconstruction for word-initial alpha with rough breathing? Is "remove commented out code" correct English? Why do I get two different answers for this counting problem? Blender 2.8 I can't see vertices, edges or faces in edit mode Where does SFDX store details about scratch orgs? 1960's book about a plague that kills all white people How can saying a song's name be a copyright violation? UK: Is there precedent for the governments e-petition site changing the ...
                    Read more

                    Schneeberg (Smreczany) Bibliografia | Menu...

                    Image
                    Szczyty w Niemczech paśmieSmreczanyn.p.m.Ochrza Schneeberg Państwo   Niemcy Położenie Bawaria Pasmo Fichtelgebirge Wysokość 1 051 m n.p.m. Położenie na mapie Bawarii Schneeberg Położenie na mapie Niemiec Schneeberg 50°03′N   11°51′E / 50,050000   11,850000 Multimedia w Wikimedia Commons Schneeberg – szczyt w paśmie Smreczany o wysokości 1051 m n.p.m., na zboczach którego znajdują się źródła rzeki Ochrza. Bibliografia | Rainer H. Schmeisser: Der Schneeberg, Beiträge zur Geschichts- und Landeskunde des Fichtelgebirges Nr. 1, Regensburg 1979 This page is only for reference, If you need detailed information, please check here
                    Read more

                    Hans Bellmer Spis treści Życiorys | Upamiętnienie | Przypisy | Bibliografia | Linki zewnętrzne |...

                    Image
                    Ludzie związani z III Liceum Ogólnokształcącym im. Adama Mickiewicza w KatowicachNiemieccy rzeźbiarzeNiemieccy malarze współcześniNiemieccy graficyLudzie urodzeni w KatowicachUrodzeni w 1902Zmarli w 1975 13 marca1902Katowicach23 lutego1975Paryżumalarzrzeźbiarzgrafikul. A. Mickiewiczaul. T. KościuszkiWyższej Szkole TechnicznejinstalacjiPaula EluardaBerlinie„tableaux vivants”NiemczechnazistówzdegenerowanąBerlinsurrealistówII wojnie światowejCamp des MillesAix-en-ProvenceParyżuUnicę ZürnPère-Lachaiseul. T. KościuszkiDział Grafiki im. Pawła Stellera Muzeum Historii Katowic2008Muzeum Historii KatowicNatalia KruszynaBerlinie Tablica pamiątkowa z 2008 roku na domu, w którym mieszkał artysta przy ul. T. Kościuszki 47 w Katowicach (2017) Grób Bellmera na cmentarzu Père-Lachaise Tablica pamiątkowa przy Ehrenfelsstraße 8a w Berlinie Budynek, w którym mieszkał artysta przy ul. T. Kościuszki 47 w Katowicach (2017) Hans Bellmer (ur. 13 marca 1902 w Katowicach [1] , zm. 23 lut...
                    Read more