Prove that the line joining mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.

In the given figure, $BE\perp AC$. $AD$ is any lire from $A$ to $BC$ intersecting $BE$ at $H. P, Q$ and $R$ are mid-points of $AH, AB$ and $BC$ respectively, then prove that $\angle PQR=90°$.

The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is $60°$. Find the angles of the parallelogram.

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

In the given figure, $ABCD$ is a parallelogram in which $P$ is the mid-point of $DC$ and $Q$ is a point on $AC$, such that $CQ=\frac{1}{4}AC$. Also, $PQ$ when produced meets $BC$ at $R$. Prove that $R$ is the mid-point of $BC$.

In the given figure, $AX$ and $CY$ are respectively the bisectors of the opposite $\angle A$ and $\angle C$ of a parallelogram $ABCD$. Show that $AX\parallel CY$.

In the given figure, $P$ is the mid-point of side $BC$ of a parallelogram $ABCD$ such that $\angle BAP=\angle DAP$. Prove that $AD=2 CD$.