Prove that the sum of the length of the diagonals of a quadrilateral is less than the perimeter, but greater than the half of the perimeter of this quadrilateral. using statement reason format.

See attached image.

(1) Prove that sum of lengths of the diagonals A and B is less that the perimeter. I am going to call A and B the full diagonal (A=A1+A2, B=B1+B2) for this part:

Use triangle inequalities:

A < a+b

A < c + d

B < c + b

B < a + d

add them up

A + A + B + B < a+b+c+d+c+b+a+d

2(A + B) < 2(a+b+c+d)

(sum of doagonals) A+B < a+b+c+d  (perimeter)   Q.E.D

(2) Prove that sum of diagonals is greater than half perimeter:

use triangle inequality:

A1+B1>a

B1+A2>b

A2+B2>c

B2+A1>d

add them up

2(A1+A2+B1+B2) > a+b+c+d

(sum of diagonals) A+B > 0.5(a+b+c+d)  (half perimeter)    Q.E.D

answer: sine

step-by-step explanation: