Question

In the given figure, ∆ABC and ∆ABD are such that AD = BC, ∠1 = ∠2 and ∠3 = ∠4. Prove that BD = AC.
Figure

Answer 1

$$\begin{gathered} \mathrm{In}∆ABC\mathrm{and}∆ABD, \\ AD\hspace{0.17em}=BC\left(\mathrm{Given}\right) \\ AB=AB\left(\text{C}\mathrm{ommon}\mathrm{side}\right) \\ \mathrm{Given}:\angle 1=\angle 2 \\ \mathrm{and}\angle 3=\angle 4 \\ \mathrm{Therefore},\angle 1+\angle 3=\angle 2+\angle 4 \\ \Rightarrow \angle DAB=\angle ABC \\ \therefore \mathrm{By}SAS,∆DAB\cong ∆CBA \\ \mathrm{Therefore},BD=AC\left(\text{C}\mathrm{orresponding}\mathrm{sides}\mathrm{of}\mathrm{congruent}\mathrm{triangles}\mathrm{are}\mathrm{equal}\right) \end{gathered}$$