Question

ABCD is quadrilateral in which $\mathrm{AB}\parallel \mathrm{CD}$. If $\mathrm{AD}=\mathrm{BC}=5\mathrm{cm}$, if $\angle \mathrm{A}=105°$, then find $\angle \mathrm{B}$.

Answer 1

Step $1:$ Drawing the diagram:

Given : $\angle \mathrm{A}=105°$

ABCD is quadrilateral in which $\mathrm{AB}\parallel \mathrm{CD}$ and $\mathrm{AD}=\mathrm{BC}=5\mathrm{cm}$.

Draw a line through C parallel to DA intersecting AB produced at E.

Step $2:$ Finding $\angle \mathrm{B}$:

We have, $\mathrm{AB}\parallel \mathrm{CD}$ and

$\mathrm{AD}\parallel \mathrm{CE}$ (By construction).

Therefore, AECD is a parallelogram.

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

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

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

$\Rightarrow \angle \mathrm{CBE}=\angle \mathrm{CEB}$ (Angles opposite to equal sides are equal)

Also, $\angle \mathrm{A}+\angle \mathrm{CBE}=180°$ (Interior angles on the same side of transversal is supplemenatry)

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

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

Hence, $\angle \mathrm{B}=105°$