In Fig., prove that:

Solution:

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 ΔABC, we have

AB + BC > AC  .... (i)

In ΔADC, we have

CD + DA > AC .... (ii)

Adding (i) and (ii), we get

AB + BC + CD + DA > AC + AC

AB + BC + CD + DA > 2AC

(ii) Now, In Δ ABC, we have,

AB + AC > BC ... (iii)

And in ΔADC, we have

CD + DA > AC

Add AB on both sides

CD + DA + AB > AC + AB

From equation (iii) and (iv), we get,

CD + DA + AB > AC + AB > BC

CD + DA + AB > BC

Hence proved