bonanova Posted March 9, 2015 Report Share Posted March 9, 2015 Consider the quadrilateral ABCD where AC and BD are its two diagonals and DA is its shortest side. UHP215 Prove (or disprove): AB + BC + CD > AC + BD Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted March 9, 2015 Report Share Posted March 9, 2015 (edited) Consider the quadrilateral ABCD where AC and BD are its two diagonals and DA is its shortest side. UHP215 Prove (or disprove): AB + BC + CD > AC + BD Suppose we have two triangles ABD and DAC. Let AB =4, BD = 5, AD = 3 Let DA = 3, AC = 4, DC = 5 Connect a line segment from BC, denote its length as x We now have a quadrilateral of ABCD AD is the shortest length of the quadrilateral AB + BC + CD > AC + BD since CD > AD now to prove a general case... Edited March 9, 2015 by BMAD Quote Link to comment Share on other sites More sharing options...
0 BMAD Posted March 10, 2015 Report Share Posted March 10, 2015 I've only been able to show it is true for cyclic quadrilaterals Quote Link to comment Share on other sites More sharing options...
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bonanova
Consider the quadrilateral ABCD where AC and BD are its two diagonals
and DA is its shortest side.
UHP215
Prove (or disprove): AB + BC + CD > AC + BD
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