O P Malhotra, S K Gupta and, Anubhuti Gangal Solutions for Exercise 2: Exercise

In figure below, $ABCD$ is a $\parallel \mathrm{gm}$. $BM$ bisects $\angle ABC$, and $DN$ bisects $\angle ADC$. Prove that: $BM=DN$

In the figure, $BM$ bisects $\angle B$ and $AN$ bisects $\angle A$ of $\parallel \mathrm{gm} ABCD$. Prove that: $ABNM$ is a rhombus.

A transversal cuts two parallel lines at $A$ and $B$. The two interior angles at $A$ are bisected and so are the two interior angles at $B$; the four bisectors from a quadrilateral $ACBD$. Prove that: $\mathrm{CD}$ is parallel to the original parallel lines.

Prove that the bisectors of any two adjacent angles of a parallelogram are at right angles.

In figure, $\mathrm{PQRS}$ is a parallelogram with bisectors $\mathrm{PA}, \mathrm{QD}, \mathrm{RC}$ and $\mathrm{SB}$ respectively of angles $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ and $\mathrm{S}$. Show that $\mathrm{ABCD}$ is a rectangle.

In the figure, $\mathrm{ABCD}$ is a parallelogram and $\mathrm{X}$ is the midpoint of $\mathrm{BC}$. The line $\mathrm{AX}$ produced meets $\mathrm{DC}$ produced at $\mathrm{Q}$. The parallelogram is  $\mathrm{ABPQ}$ completed. Prove that: $∆\mathrm{ABX}\cong ∆\mathrm{QCX}$

In the figure, $\mathrm{ABCD}$ is a parallelogram and $\mathrm{X}$ is the midpoint of $\mathrm{BC}$. The line $\mathrm{AX}$ produced meets $\mathrm{DC}$ produced at $\mathrm{Q}$. The parallelogram $\mathrm{ABPQ}$ is completed. Prove that: $\mathrm{DC}=\mathrm{CQ}=\mathrm{QP}$

$\mathrm{ABCD}$ is a rhombus with $\mathrm{P}, \mathrm{Q}, \mathrm{R}$ as midpoints of $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{CD}$. Prove that $\mathrm{PQ}\perp \mathrm{QR}$.