R S Aggarwal and Veena Aggarwal Solutions for Exercise 1: Exercise 7A

In a $\Delta ABC, AD$ is the bisector of $\angle A$.
If $AB=10 \mathrm{cm}, AC=14 \mathrm{cm}$ and $BC=6 \mathrm{cm}$, find $BD$ and $DC$.

$M$ is a point on the side $BC$ of a parallelogram  $ABCD$. $DM$ when produced meets $AB$ produced at $N$ Prove that
$\frac{DN}{DM}=\frac{AN}{DC}$

In $\Delta ABC$ , $M$ and $N$ are points on the sides $AB$ and $AC$ respectively such that $BM=CN$. If $\angle B=\angle C$ then show that $MN\left|\right|BC$.

$∆ABC$ and $\Delta DBC$ lie on the same side of $BC,$ as shown in the figure. From a point $P$ on $BC, PQ \left|\right| AB$ and $PR \left|\right| BD$ are drawn, meeting $AC$ at $Q$ and $CD$ at $R,$ respectively. Prove that $QR \left|\right| AD.$

In the given figure, side $BC$ of $∆ABC$ is bisected at $D$ and $O$ is any point on $AD. BO$ and $\mathrm{CO}$ produced meet $A\mathrm{C}$ and $AB$ at $E$ and $F$ respectively, and $AD$ is produced to $X$ so that $D$ is the midpoint of $OX.$ Prove that $AO:AX=AF:AB$ and show that $EF\parallel BC.$

$ABCD$ is a parallelogram in which $P$ is the midpoint of $DC$ and $Q$ is a point on $AC$ such that $CQ=\frac{1}{4}AC$. If $PQ$ produced meets $BC$ at $R$, prove that $R$ is the midpoint of $BC$.

In the adjoining figure, $\mathrm{ABC}$ is a triangle in which $\mathrm{AB}=\mathrm{AC}$. If $\mathrm{D}$ and $\mathrm{E}$ are points on $\mathrm{AB}$ and $\mathrm{AC}$ respectively such that $\mathrm{AD}=\mathrm{AE}$, show that the points $\mathrm{B}, \mathrm{C}, \mathrm{E}$ and $\mathrm{D}$ are concyclic.

In $\mathrm{\Delta ABC}$, the bisector of $\angle \mathrm{B}$ meets $\mathrm{AC}$ at $\mathrm{D}$. A line $\mathrm{PQ}\parallel \mathrm{AC}$ meets $\mathrm{AB}, \mathrm{BC}$ and $\mathrm{BD}$ at $P$, $\mathrm{Q}$ and $\mathrm{R}$ respectively. Show that $\mathrm{BP}\times \mathrm{QR}=\mathrm{BQ}\times \mathrm{PR}$.