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Understanding Ceva's Theorem

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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 $fracA_triangle PXBA_triangle PXC=fracBXCX=fracA_triangle BXAA_triangle CXA$.



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



$fracA_triangle APBA_triangle APC=fracA_triangle AXB - A_triangle PXBA_triangle AXC-A_triangle PXC=fracBXCX$



How does the subtraction of the two areas make it so that the new triangles are still proportional to $fracBXCX$? (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 $fracA_triangle PXBA_triangle PXC=fracBXCX=fracA_triangle BXAA_triangle CXA$.



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



    $fracA_triangle APBA_triangle APC=fracA_triangle AXB - A_triangle PXBA_triangle AXC-A_triangle PXC=fracBXCX$



    How does the subtraction of the two areas make it so that the new triangles are still proportional to $fracBXCX$? (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 $fracA_triangle PXBA_triangle PXC=fracBXCX=fracA_triangle BXAA_triangle CXA$.



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



      $fracA_triangle APBA_triangle APC=fracA_triangle AXB - A_triangle PXBA_triangle AXC-A_triangle PXC=fracBXCX$



      How does the subtraction of the two areas make it so that the new triangles are still proportional to $fracBXCX$? (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 $fracA_triangle PXBA_triangle PXC=fracBXCX=fracA_triangle BXAA_triangle CXA$.



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



      $fracA_triangle APBA_triangle APC=fracA_triangle AXB - A_triangle PXBA_triangle AXC-A_triangle PXC=fracBXCX$



      How does the subtraction of the two areas make it so that the new triangles are still proportional to $fracBXCX$? (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 2 hours ago









      YuiTo Cheng

      2,48341037




      2,48341037










      asked 2 hours ago









      dragonkingdragonking

      384




      384




















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          $A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$



          $A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$



          Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            1 Answer
            1






            active

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            active

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            active

            oldest

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            3












            $begingroup$

            $A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$



            $A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$



            Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$






            share|cite|improve this answer









            $endgroup$

















              3












              $begingroup$

              $A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$



              $A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$



              Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$






              share|cite|improve this answer









              $endgroup$















                3












                3








                3





                $begingroup$

                $A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$



                $A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$



                Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$






                share|cite|improve this answer









                $endgroup$



                $A_triangle AXB: A_triangle AXC=BX:CXRightarrow A_triangle AXB=fracBXCXA_triangle AXC$



                $A_triangle PXB: A_triangle PXC=BX:CXRightarrow A_triangle PXB=fracBXCXA_triangle PXC$



                Hence $$fracA_triangle A X B -A _triangle P X BA _triangle A X C-A_ triangle P X C=fracfracBXCXA_triangle AXC-fracBXCXA_triangle PXCA _triangle A X C-A_ triangle P X C=fracBXCX$$







                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



























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