Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)
Solution:
Given, ABCD is a quadrilateral
We have to show that AB + BC + CD + DA < 2 (BD + AC)
Join the diagonals AC and BD of the quadrilateral.
Consider the triangle OAB,
We know that the sum of two sides of a triangle is greater than the third side.
OA + OB > AB ------------ (1)
Consider the triangle OBC,
We know that the sum of two sides of a triangle is greater than the third side.
OB + OC > BC ------------ (2)
Consider the triangle OCD,
OC + OD > CD ------------ (3)
In triangle ODA,
OD + OA > DA -------------- (4)
Adding (1), (2), (3) and (4), we get
OA + OB + OB + OC + OC + OD + OD + OA > AB + BC + CD + DA
2(OA + OB + OC + OD) > AB + BC + CD + DA
From the figure,
AC = OA + OC
BD = OB + OD
2[(OA + OC) + (OB + OD)] > AB + BC + CD + DA
2(AC + BD) > AB + BC + CD + DA
The above expression can be rewritten as
AB + BC + CD + DA < 2(AC + BD)
Therefore, it is proven that AB + BC + CD + DA < 2(BD + AC)
✦ Try This: In the figure, it is given that AE = AD and BD = CE. Prove that △AEB is congruent △ADC.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 11
Show that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)
Summary:
It is shown that in a quadrilateral ABCD, AB + BC + CD + DA < 2 (BD + AC)
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