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

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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?












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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 2 hours ago









      YuiTo Cheng

      2,48341037




      2,48341037










      asked 2 hours ago









      dragonkingdragonking

      384




      384






















          1 Answer
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          $begingroup$

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



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



          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









          $endgroup$














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            $begingroup$

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



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



            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






























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