Question

Prove that the opposite angle of an isosceles trapezium are supplementary.

Answer 1

Given: In trapezium ABCD, AB = CD and AD || BC.

Construction: Draw AE$\perp$BC and DF$\perp$BC.

To prove: $\angle \mathrm{A}+\angle \mathrm{C}=180°\mathrm{and}\angle \mathrm{B}+\angle \mathrm{D}=180°$

Proof:

$$\begin{gathered} \mathrm{In}∆\mathrm{ABE}\mathrm{and}∆\mathrm{DCF}, \\ \mathrm{AB}=\mathrm{CD}\left(\mathrm{Given}\right) \\ \mathrm{AE}=\mathrm{DF}\left(\mathrm{Perpendicular}\mathrm{distance}\mathrm{between}\mathrm{the}\mathrm{parallel}\mathrm{lines}\right) \\ \angle \mathrm{AEB}=\angle \mathrm{DFC}=90° \\ \therefore \mathrm{By}\mathrm{RHS}\mathrm{congruence}\mathrm{criterion} \\ ∆\mathrm{ABE}\cong \mathrm{DCF} \\ \Rightarrow \angle \mathrm{ABE}=\angle \mathrm{DCF}\left(\mathrm{By}\mathrm{CPCT}\right) \\ \Rightarrow \angle \mathrm{ABC}=\angle \mathrm{DCB}.....\left(\mathrm{i}\right) \\ \mathrm{Now},\mathrm{AD}\parallel \mathrm{BC}\mathrm{and}\mathrm{AB}\mathrm{is}\mathrm{a}\mathrm{transversal} \\ \Rightarrow \angle \mathrm{ABC}+\angle \mathrm{BAD}=180°\left(\mathrm{Consecutive}\mathrm{interior}\mathrm{angles}\right) \\ \Rightarrow \angle \mathrm{DCB}+\angle \mathrm{BAD}=180°\left[\mathrm{using}\left(\mathrm{i}\right)\right] \\ \therefore \angle \mathrm{C}+\angle \mathrm{A}=180° \\ \mathrm{Also},\mathrm{AD}\parallel \mathrm{BC}\mathrm{and}\mathrm{CD}\mathrm{is}\mathrm{a}\mathrm{transversal} \\ \Rightarrow \angle \mathrm{ADC}+\angle \mathrm{DCB}=180°\left(\mathrm{Consecutive}\mathrm{interior}\mathrm{angles}\right) \\ \Rightarrow \angle \mathrm{ADC}+\angle \mathrm{ABC}=180°\left[\mathrm{using}\left(\mathrm{i}\right)\right] \\ \therefore \angle \mathrm{D}+\angle \mathrm{B}=180° \end{gathered}$$

Hence, the opposite angle of an isosceles trapezium are supplementary.