Question

In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD > BC.

Answer 1

Given: ∠B < ∠A and ∠C < ∠D

To prove: AD > BC

Proof:

$$\begin{gathered} \mathrm{In}∆AOB, \\ \angle B<\angle A \\ \Rightarrow AO<BO\left(\mathrm{Side}\mathrm{opposite}\mathrm{to}\mathrm{the}\mathrm{greater}\mathrm{angle}\mathrm{is}\mathrm{longer}\right).....\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In}∆COD, \\ \angle C<\angle D \\ \Rightarrow OD<OC\left(\mathrm{Side}\mathrm{opposite}\mathrm{to}\mathrm{the}\mathrm{greater}\mathrm{angle}\mathrm{is}\mathrm{longer}\right).....\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \\ AO+OD<BO+OC \\ \therefore AD<BC \end{gathered}$$