Question

In the given figure, prove that
(i) CD + DA + AB > BC
(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)
$$\begin{gathered} \mathrm{In}∆ACD, \\ CD+DA>CA...\left(1\right) \\ \mathrm{In}∆ABC, \\ AB+CA>BC...\left(2\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ CD+DA+AB+CA>CA+BC \\ \therefore AB+CD+DA>BC \end{gathered}$$

(ii)
$$\begin{gathered} \mathrm{In}∆CDA, \\ CD+DA>CA...\left(3\right) \\ \mathrm{In}∆BCA, \\ BC+AB>CA...\left(4\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Adding}\left(3\right)\mathrm{and}\left(4\right),\mathrm{we}\mathrm{get} \\ CD+AD+BC+AB>CA+CA \\ \therefore CD+AD+BC+AB>2CA \end{gathered}$$