In Fig. 10.131, prove that:


(i) CD + DA + AB + BC > 2AC


(ii) CD+DA +AB > BC

Given to prove,

(i) CD + DA + AB + BC > 2AC


(ii) CD+DA +AB > BC


From the given figure,


We know that,


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


(i) So,


In , we have


AB + BC > AC (1)


In , we have


CD + DA > AC (2)


Adding (1) and (2), we get


AB + BC + CD + DA > AC + AC


CD + DA + AB + BC > 2 AC


(ii) Now, in , we have


CD + DA > AC


Add AB on both sides, we get


CD + DA +AB > AC + AB > BC


CD + DA + AB > BC


Hence, proved


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