Question

Given: $\mathrm{ABCX}$is a is a localid="1637648903053" $▱\mathbf{,}\mathbf{}\mathrm{DXFE}$is a $▱$.

Prove: $\angle \mathrm{B}\cong \angle \mathrm{E}$

Short answer

It is proved that $\angle \mathrm{B}\cong \angle \mathrm{E}$.

Step-by-step solution

Step 1. Property of parallelogram.

The opposite angles of a parallelogram are congruent.

It is given that ABCX is a parallelogram in which $\angle B$and $\angle AXC$are opposite angles of a parallelogram. Then by property of parallelogram, $\angle B\cong \angle AXC$.

Step 2. Vertical opposite angles.

From the figure, it can be observed that $\angle AXC$and $\angle DXF$are vertically opposite angles. Therefore, by the definition of vertically opposite angles, $\angle AXC\cong \angle DXF$.

Step 3. Property of parallelogram.

It is given that DEFX is a parallelogram in which $\angle E$and $\angle DXF$are opposite angles of a parallelogram. Then by property of parallelogram, $\angle DXF\cong \angle E$.

As $\angle B\cong \angle AXC$, $\angle AXC\cong \angle DXF$and $\angle DXF\cong \angle E$. It can be concluded that $\angle B\cong \angle E$.