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Pls tell all the answers given in this image

Dear student,

O is a point in the interior of a given triangle, then three triangles ΔOPQ, ΔOQR, and ΔORP can be constructed. In a triangle, the sum of the lengths of either two sides is always greater than the third side.

(i) Yes, as ΔOPQ is a triangle with sides OP, OQ, and PQ.

OP + OQ > PQ

(ii) Yes, as ΔOQR is a triangle with sides OR, OQ, and QR.

OQ + OR > QR

(iii) Yes, as ΔORP is a triangle with sides OR, OP, and PR.

OR + OP > PR

3 We Know that,
In a triangle sum of two sides is greater than the third side
Hence,In triangle ABM 
AB + BM > AM  .........1
Also in triangle ACM
AC + MC > AM ..........2
Adding equation 1 and 2 we get,
AB + BM + AC + MC  > AM + AM
AB + AC + (BM + MC) > 2AM
AB + AC + BC > 2AM

regrads 
Tanmayee

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5)AC is Hypotenuse in triangle ABC.
so,
AB<AC---1
BC<AC---2
because hypotenuse is the biggest one in Right triangle.And ABCD is quardrilateral so ABC is Right triangle.
similarly,
DA<BD---3
CD<BD---4
Now from 1,2,3& 4
AB+BC+CD+DA<(AC+AC)+(BD+BD)
=>AB+BC+CD+DA<2(AC+BD)

4)

Num of the two sides of a triangle is greater than the third side.
So, considering the triangle ABC, BCD, CAD and BAD, we get
AB + BC > AC
CD + AD > AC
AB + AD > BD
BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)
⇒ 2(AB + BC + CA + AD) > 2(AC + BD)
⇒ (AB + BC + CA + AD) > (AC + BD)
 

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