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