Question

In the given figure, ABCD is a quadrilateral in which AB || DC and AD || BC.
Prove that ∠ADC = ∠ABC.

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{AB}\parallel \mathrm{CD} \\ \mathrm{AD}\parallel \mathrm{BC} \\ \angle 1+\angle 2=180°(\mathrm{AB}\parallel \mathrm{CD}\mathrm{and}\mathrm{AD}\mathrm{is}\mathrm{the}\mathrm{transversal})...\left(\mathrm{i}\right) \\ \angle 2+\angle 3=180°(\mathrm{AD}\parallel \mathrm{BC}\mathrm{and}\mathrm{AB}\mathrm{is}\mathrm{the}\mathrm{transversal})...\left(\mathrm{ii}\right) \\ \mathrm{From}\left(\mathrm{i}\right)\mathrm{and}\left(\mathrm{ii}\right): \\ \angle 1+\angle 2=180°=\angle 2+\angle 3 \\ \angle 1=\angle 3 \\ \angle \mathrm{ADC}=\angle \mathrm{ABC} \end{gathered}$$