Question

Let ABCD be a quadrilateral with diagonals AC and BD. Prove the following statements (Compare these with the previous problem);

(a) AB + BC + CD > AD;

(b) AB + BC + CD + DA > 2AC;

(c) AB + BC + CD + DA > 2BD;

(d) AB + BC + CD + DA > AC + BD.

Answer 1

(a) In ΔABC:

AB + BC > AC (The sum of any two sides of a triangle is greater than the third side)

Adding CD to both sides, we get:

AB + BC + CD > AC + CD … (1)

In ΔACD:

The sum of any two sides of a triangle is greater than the third side.

AC + CD > AD … (2)

From (1) and (2), we get:

AB + BC + CD > AD

(b) In ΔABC:

The sum of any two sides of a triangle is greater than the third side.

AB + BC > AC ... (3)

In ΔACD:

CD + DA > AC … (4)

Adding (3) and (4), we get:

AB + BC + CD + DA > AC + AC

AB + BC + CD + DA > 2AC

(c) In ΔABD:

The sum of any two sides of a triangle is greater than the third side.

AB + AD > BD ... (3)

In ΔBCD:

BC + CD > BD … (4)

Adding (3) and (4), we get:

AB + AD + BC + CD > BD + BD

AB + BC + CD + DA > 2BD

(d) We know that in any triangle, the sum of two sides is always greater than the third side.

AB + BC > AC

BC + CD > BD

CD + AD > AC

AD + AB > BD

Adding all the inequalities, we get:

2 (AB + BC + CD + AD) > 2 (AC + BD)

AB + BC + CD + AD > AC + BD