Question

In the given figure, in ∆ABC, it is given that ∠B = 40° and ∠C = 50°. DE || BC and EF || AB. Find: (i) ∠ADE + ∠MEN (ii) ∠BDE and (iii) ∠BFE.
Figure

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)DE\parallel BC \\ \mathrm{Therefore}, \\ \angle ADE=\angle ABC=40°\left(\text{A}\mathrm{lternate}\mathrm{angle}s\right) \\ EF\parallel AB\text{and}MF\mathrm{is}\text{the}\mathrm{produced}\mathrm{line}\mathrm{of}EF. \\ \mathrm{Therefore},MF\parallel AB. \\ \Rightarrow \angle MEN=\angle ADE=40°\left(\text{A}\mathrm{lternate}\mathrm{angle}s\right) \\ \angle ADE+\angle MEN=40°+40°=80° \\ \left(\mathrm{ii}\right)DE\parallel BC\mathrm{and}AB\mathrm{is}\text{the}\mathrm{transversal}. \\ \mathrm{Interior}\mathrm{angles}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\text{a}\mathrm{transversal}\mathrm{are}\mathrm{supplementary}. \\ \therefore \angle BDE=180°-\angle ABC \\ =180°-40° \\ =140° \\ \left(\mathrm{iii}\right)EF\parallel AB\mathrm{and}BC\mathrm{is}\text{the}\mathrm{transversal}. \\ \mathrm{Interior}\mathrm{angles}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\text{a}\mathrm{transversal}\mathrm{are}\mathrm{supplementary}. \\ \therefore \angle BFE=180°-\angle ABC \\ =180°-40° \\ =140° \end{gathered}$$