In any quadrilateral, the sum of the lengths of its four sides exceeds the sum of the lengths of its diagonals. Prove it.

Let ABCD be a quadrilateral.

Construction: Join BD and AC as its diagonals.

To Prove: In any quadrilateral, the sum of the lengths of its four sides is greater than the sum of the lengths of its diagonals.

Proof: Using the Triangle Inequality Theorem, which states that the sum of the lengths of any two sides of a triangle is always greater than the length of the third side:

In Δ ABC:

AB + CB > AC ……………….(1)

Similarly, in Δ BCD:

BC + CD > BD ……………….(2)

Similarly, in Δ CDA:

CD + DA > AC ……………….(3)

Similarly, in Δ DAB:

DA + AB > BD ……………….(4)

Now, adding the inequalities (1) and (2),

(AB + BC) + (BC + CD) > AC + BD

⇒ AB + BC + BC + CD > AC + BD

⇒ AB + 2BC + CD > AC + BD ………………..(5)

Adding the inequalities from (3) and (4),

(CD + DA) + (DA + AB) > AC + BD

⇒ CD + DA + DA + AB > AC + BD

⇒ CD + 2DA + AB > AC + BD ……………….(6)

Adding equations (5) and (6), we get:

⇒ (AB + 2BC + CD) + (CD + 2DA + AB) > (AC + BD) + (AC + BD)

⇒ AB + 2BC + CD + CD + 2DA + AB > 2(AC + BD)

⇒ 2(AB + BC + CD + DA) > 2(AC + BD)

⇒ AB + BC + CD + DA > AC + BD

Hence, the sum of the lengths of the four sides of a quadrilateral is greater than the sum of the lengths of its diagonals.