The problem describes a quadrilateral formed from a parallelogram and a triangle. a) We need to identify the type of quadrilateral. b) We need to calculate the size of the angle $DFE$.

1. Problem Description

The problem describes a quadrilateral formed from a parallelogram and a triangle.

a) We need to identify the type of quadrilateral.

b) We need to calculate the size of the angle $DFE$.

2. Solution Steps

a) The quadrilateral has one pair of parallel sides ($CD$ and $GF$). A quadrilateral with one pair of parallel sides is called a trapezium (or trapezoid).

b) Since $CDFG$ is a parallelogram, opposite angles are equal, and adjacent angles are supplementary (add up to $180^\circ$). Therefore, angle $CGF$ is supplementary to angle $DCG$.

$\angle CGF = 180^\circ - 116^\circ = 64^\circ$.

Since $CDFG$ is a parallelogram, $CD$ is parallel to $GF$, so $FD$ is parallel to $CG$. Therefore, triangle $DFE$ is isosceles with $DF=DE$.

Also, in a parallelogram, opposite sides are parallel and equal in length. So, $CD \parallel GF$ and $CG \parallel DF$. Since $CG \parallel DF$ and $DF = DE$, triangle $DFE$ is an isosceles triangle. In an isosceles triangle, the angles opposite to the equal sides are equal. Therefore, $\angle DFE = \angle DEF$.

In parallelogram $CDFG$, $\angle CFD = \angle CGF = 64^\circ$.

Since $DF=DE$, the triangle $DFE$ is an isosceles triangle. Therefore, $\angle DFE = \angle DEF$.

The sum of angles in a triangle is $180^\circ$.

$\angle DFE + \angle DEF + \angle EDF = 180^\circ$.

Since $\angle DFE = \angle DEF$,

$2 \times \angle DFE + \angle EDF = 180^\circ$.

Since $CD \parallel GF$, $\angle CDF = \angle CGF = 64^\circ$.

Also, $CDFG$ is a parallelogram, so $\angle G = \angle D$. This means $\angle G = \angle CDF = 64^\circ$.

Also, because DF = DE, Triangle DFE is isosceles and angles DFE and DEF are equal.

The angles of Triangle DFE should sum to

1

8

0. Since $\angle CFD$ and $\angle DFE$ are angles on a straight line:

$64^\circ + \angle DFE + \angle CDE = 180^\circ$.

$\angle DFE + \angle DEF + \angle EDF = 180^\circ$

$\angle DFE = \angle DEF$ because $DF=DE$

Then, $2 \angle DFE + \angle EDF = 180^\circ$

$\angle EDF = 180 - (2 \times \angle DFE)$

Also, since $CD \parallel GF$, $\angle CDF = \angle CGF = 64^\circ$.

$CDFG$ is a parallelogram, so opposite angles are equal. Thus, $\angle DGF = 180^\circ - 116^\circ = 64^\circ$. Also, since sides $DF$ and $DE$ are of equal length, $\angle DFE = \angle DEF$. If we say that $\angle DFE = x$, then, the $\angle FDE = 180 - 2x$

Sum of the angles on line $CFE$ equal $180^\circ$

$\angle CFD + \angle DFE = 180^\circ$

Also given is that $CD \parallel FE$. Therefore, $\angle CDF = \angle CGF = 64^\circ$. And since $DF = DE$, $\angle DFE = \angle DEF$.

Then the angles in the triangle $DFE$ sum up to 180 degrees.

$\angle EDF + \angle DFE + \angle DEF = 180^\circ$

$\angle EDF + 2 \angle DFE = 180^\circ$

In our image, since DF = DE, triangle DFE is isosceles. Since $CD \parallel GF$, $\angle CDF = 64$. Because DF = DE, we have $\angle DFE = \angle DEF$. Thus $(180-64) / 2 = 116/2 = 58$ because triangle DFE must be isosceles

3. Final Answer

a) Trapezium

b) 58°