Question

$\square \mathrm{ABCD}$ is a trapezium in which $\mathrm{AB}\left|\right|\mathrm{CD}$ and$\mathrm{AD}=\mathrm{BC}$(see Fig.) .

Show that $\angle \mathrm{A}=\angle \mathrm{B}$ ( [Hint : Extend $\mathrm{AB}$ and draw a line through $\mathrm{C}$parallel to $\mathrm{DA}$ intersecting $\mathrm{AB}$ produced at $\mathrm{E}$.]

Answer 1

Showing that the $\angle \mathrm{A}=\angle \mathrm{B}$:

Given:

The trapezium $\mathrm{ABCD}$ in which

$\mathrm{AB}\left|\right|\mathrm{CD}$

$\mathrm{AD}=\mathrm{BC}$ (non parallel sides of trapezium are equal)

Construction:

Extended $\mathrm{AB}$ and draw a line through $\mathrm{C}$ parallel to $\mathrm{DA}$ intersecting $\mathrm{AB}$produced at$\mathrm{E}$

$▱\mathrm{ADCE}$ become parallelogram

Solution:.

In

$\mathrm{CE}=\mathrm{AD}$ (Opposite sides of a parallelogram)

$\mathrm{AD}=\mathrm{BC}$(Given)

$\mathrm{BC}=\mathrm{CE}$

$\Rightarrow \angle \mathrm{CBE}=\angle \mathrm{CEB}$ (opposite angles of the equal sides)

also,

$\mathrm{AD}\parallel \mathrm{CE}$

$\angle \mathrm{A}+\angle \mathrm{CEB}=180°$(co-interior angle)

$\angle \mathrm{A}+\angle \mathrm{CBE}=180°$ ( $\angle \mathrm{CBE}=\angle \mathrm{CEB}$)

$\angle \mathrm{A}=180°-\angle \mathrm{CBE}$------------------------($1$)

$\angle \mathrm{B}+\angle \mathrm{CBE}=180°$( As Linear pair)

$\angle \mathrm{B}=180°-\angle \mathrm{CBE}$-----------------------($2$)

From $1$ and $2$

$\Rightarrow \angle \mathrm{A}=\angle \mathrm{B}$

Hence the $\angle \mathrm{A}=\angle \mathrm{B}$