Question

$\text{In the given figure, prove that}\text{(i) CD + DA + AB}>BC\text{(ii) CD + DA + AB + BC}>2AC.$

Answer 1

Given: Quadrilateral ABCD
To prove:
(i) CD + DA + AB > BC
(ii) CD + DA + AB + BC > 2AC
Proof:
(i)
In ∆ACD,CD+DA>CA
...1In ∆ABC,AB+CA>BC
...2
Adding (1) and (2), we getCD+DA+AB+CA>CA+BC∴ AB+CD+DA>BC
(ii)
In ∆CDA,CD+DA>CA
...3In ∆BCA,BC + AB > CA
...4
Adding (3) and (4), we getCD+AD+BC+AB>CA+CA∴ CD+AD+BC+AB>2CA