Question

Prove that in a cyclic trapezium angles at the base are congruent.

Answer 1

Given: ABCD is a cyclic trapezium.
To prove: $\angle \mathrm{C}=\angle \mathrm{D}$
Proof:

$$\begin{gathered} \angle \mathrm{C}+\angle \mathrm{A}=180°...\left(\mathrm{i}\right)(\mathrm{ABCD}\mathrm{is}\mathrm{a}\mathrm{cyclic}\mathrm{quadrilateral}) \\ \angle \mathrm{A}+\angle \mathrm{D}=180°...\left(\mathrm{ii}\right)(\mathrm{Angles}\mathrm{on}\mathrm{the}\mathrm{same}\mathrm{side}\mathrm{of}\mathrm{a}\mathrm{transversal}\mathrm{line}\mathrm{are}\mathrm{supplementary}) \\ \mathrm{Subtracting}\mathrm{equaltion}\left(\mathrm{ii}\right)\mathrm{from}\left(\mathrm{i}\right),\mathrm{we}\mathrm{get}: \\ \angle \mathrm{C}+\angle \mathrm{A}-\angle \mathrm{A}-\angle \mathrm{D}=0 \\ \Rightarrow \angle \mathrm{C}=\angle \mathrm{D} \end{gathered}$$