Prove that the sum of the length of the diagonals of a quadrilateral is less than the perimeter of the quadrilateral.

Question

Note: Quadrilateral is any four-sided figure.

Diagonal is the segment that connect the opposite vertices.

Perimeter is the sum of the length of the figure.

Answer 1

AC < AB + CB and AC < AD + CD

BD < AD + AB and BD < DC + CB

This is by triangle inequality.

If we ad all 4 inequalities we get: 2AC + 2BD = 2AB + 2CB + 2AD + 2DC or

AC + BD < AB + BC + CD + AD.

This is end of prove.

Answer 2

AC > AD + DC

BD > AB + BC

AC + BC > AB + BC + CD + DA