In a parallelogram $ABCD$,$AB = 10 \, cm$ and $AD = 6 \, cm$. The bisector of $\angle A$ meets $DC$ in $E$. $AE$ and $BC$ produced meet at $F$. Find the length of $CF$ (in $cm$).

$ABCD$ is a trapezium such that $AB || CD$. If $\angle A = y + 60^{\circ}$,$\angle B = x + 60^{\circ}$,$\angle C = 3x - 40^{\circ}$ and $\angle D = 3y - 80^{\circ}$,then find the measure of each angle of $ABCD$.

The quadrilateral formed by joining the mid-points of the sides of a quadrilateral $PQRS,$ taken in order,is a rhombus,if

In parallelogram $ABCD$,$\angle A$ is an obtuse angle,$AM \perp BC$,and $AN \perp CD$. If $\angle MAN = 60^{\circ}$,find each of the angles of quadrilateral $ABCD$.

Through $A, B$ and $C$,lines $RQ, PR$ and $QP$ have been drawn,respectively parallel to sides $BC, CA$ and $AB$ of a $\triangle ABC$ as shown in the figure. Show that $BC = \frac{1}{2} QR$.

$A$ square is inscribed in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.