If the diagonals of a parallelogram are equal, then show that it is a rectangle.

 Diagonal $\mathrm{A}\mathrm{C}$ of a parallelogram $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D}$ bisects $\mathrm{\angle }\mathrm{A}$ (see Fig). Show that it bisects $\mathrm{\angle }\mathrm{C}$ also.

In parallelogram $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D},$ two points $\mathrm{P}$ and $\mathrm{Q}$ are taken on diagonal $\mathrm{B}\mathrm{D}$ such that $\mathrm{D}\mathrm{P}=\mathrm{B}\mathrm{Q}$. Show that $\mathrm{\Delta }\mathrm{A}\mathrm{P}\mathrm{D}\cong \mathrm{\Delta }\mathrm{C}\mathrm{Q}\mathrm{B}$.

In parallelogram $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D},$ two points $\mathrm{P}$ and $\mathrm{Q}$ are taken on diagonal $\mathrm{B}\mathrm{D}$ such that $\mathrm{D}\mathrm{P}=\mathrm{B}\mathrm{Q}$. Show that: $\mathrm{A}\mathrm{P}=\mathrm{C}\mathrm{Q}$

In parallelogram $\mathrm{A}\mathrm{B}\mathrm{C}\mathrm{D},$ two points $\mathrm{P}$ and $\mathrm{Q}$ are taken on diagonal $\mathrm{B}\mathrm{D}$ such that $\mathrm{D}\mathrm{P}=\mathrm{B}\mathrm{Q}$. Show that: $\mathrm{\Delta }\mathrm{A}\mathrm{Q}\mathrm{B}\cong \mathrm{\Delta }\mathrm{C}\mathrm{P}\mathrm{D}$