problem on quadrileteral

ram9 · New Member Offline Joined: 29 Oct 2009 Posts: 1

please help me solve this problem
In a quadrilateral ABCD prove that AB+BC+CD+CD<2(AC+BD) the sum of four sides is less than twice the sum of its diagonals

**Posted:** Thu Oct 29, 2009 1:24 am

rationalist · New Member Offline Joined: 04 Aug 2009 Posts: 15

use the idea that sum of two sides of a triangle is greater than the third side.
diagonals form four triangles in the quadrilateral

**Posted:** Thu Oct 29, 2009 10:12 am

AwesomeToad

Proofs should be moved to the High school Basics Forum

This forum is not for proofs

**Posted:** Thu Oct 29, 2009 10:19 am

fishythefish

Let the intersection of the diagonals be $E$ .

$AE + BE > AB$
$BE + CE > BC$
$CE + DE > CD$
$AE + DE > AD$

$AE + CE = AC$
$BE + DE = BD$

$2(AE + BE + CE + DE) > AB + BC + CD + AD$
$2(AC + BC) > AB + BC + CD + AD$

**Posted:** Thu Oct 29, 2009 12:26 pm

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