In the figure,$ABCD$ and $AEFG$ are two parallelograms. If $\angle C = 55^{\circ},$ determine $\angle F.$ (in $^{\circ}$)

$E$ and $F$ are respectively the mid-points of the non-parallel sides $AD$ and $BC$ of a trapezium $ABCD$. Prove that $EF \parallel AB$ and $EF = \frac{1}{2}(AB + CD)$.

Can all the four angles of a quadrilateral be obtuse angles? Give reason for your answer.

$E$ is the mid-point of the side $AD$ of the trapezium $ABCD$ with $AB \parallel DC$. $A$ line through $E$ drawn parallel to $AB$ intersects $BC$ at $F$. Show that $F$ is the mid-point of $BC$.

In quadrilateral $ABCD$,$AC = BD$ and $AC \perp BD$. If $P, Q, R$ and $S$ are the midpoints of the sides $AB, BC, CD$ and $DA$ respectively,prove that $PQRS$ is a square.

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.