# Show that in a qu

Here, ABCD is a quadrilateral and AC and BD are its diagonals.

Now, As we that, sum of two sides of a triangle is greater than the third side.

In Δ ACB,

AB + BC > AC (i)

In Δ BDC,

CD + BC > BD (ii)

In Δ ACD,

AD + DC > AC (iv)

Now, adding (i), (ii), (iii) and (iv):

AB + BC + CD + BC + AB + AD + AD + DC> AC + BD + BD + AC

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

Thus, AB + BC + CD + AD > AC + BD

Hence, proved

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Revision on Angle sum Property of Triangles44 mins
Concentration of a Solution - Part 246 mins
Concentration of a Solution - Part 142 mins
Variation in g with Rotation of Earth42 mins
Surface Area and Volume of solids Revision43 mins
Arc of Circle And Related IMP Qs40 mins
Circles- All kinds of Questions36 mins
Significance of Newton's Laws in daily life42 mins
Sphere and Hemisphere42 mins
Structural Organisation In Animals-I54 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

Each question conRS Aggarwal & V Aggarwal - Mathematics

In the given figuRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics

<span lang="EN-USRS Aggarwal & V Aggarwal - Mathematics