Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD
Solution:
Given, ABCD is a quadrilateral
We have to show that AB + BC + CD + DA > AC + BD
Join the diagonals AC and BD of the quadrilateral.
Considering triangle ABC,
We know that the sum of two sides of a triangle is greater than the third side.
AB + BC > AC ---------------- (1)
Considering triangle BCD,
BC + CD > BD ---------------- (2)
Considering triangle CDA,
CD + DA > AC ---------------- (3)
Considering triangle DAB,
DA + AB > BD ----------------- (4)
Adding (1), (2), (3) and (4), we get
AB + BC + BC + CD + CD + DA + DA + AB > AC + BD + AC + BD
On rearranging,
AB + AB + BC + BC + CD + CD + DA + DA > AC + AC + BD + BD
2(AB + BC + CD + DA) > 2(AC + BD)
Therefore, AB + BC + CD + DA > AC + BD
✦ Try This: In figure, AD=BD and ∠C=∠E. Prove that BC=AE.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 7
NCERT Exemplar Class 9 Maths Exercise 7.4 Problem 12
Show that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD
Summary:
It is shown that in a quadrilateral ABCD, AB + BC + CD + DA > AC + BD
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