B Nirmala Shastry Solutions for Exercise 1: EXERCISE 9

In a quadrilateral $\mathrm{ABCD}$, $\mathrm{P}$ is an internal point. Prove that $\mathrm{AP}+\mathrm{BP}+\mathrm{CP}+\mathrm{DP}>$ Semi-perimeter of quadrilateral $\mathrm{ABCD}$.

The sides $\mathrm{BA}$ and $\mathrm{BC}$ of $△\mathrm{ABC}$ are produced to $\mathrm{D}$ and $\mathrm{E}$ respectively. Prove that $\angle \mathrm{DAC}+\angle \mathrm{ECA}>{180}^{°}$.

In $△\mathrm{PQR}, \angle \mathrm{Q}={70}^{°}, \angle \mathrm{QPR}={50}^{°} \text{and } \mathrm{PR}=\mathrm{RS}. \mathrm{Prove} \mathrm{that} \mathrm{PQ}<\mathrm{PS}$.

In the quadrilateral $\mathrm{ACED}, \angle \mathrm{CAD}={100}^{°}=\angle \mathrm{D} \text{and} \angle \mathrm{E}={50}^{°}$. Show that $\mathrm{DE}<\mathrm{DB}$.

In the quadrilateral $\mathrm{ACED}, \angle \mathrm{CAD}={100}^{°}=\angle \mathrm{D} \text{and} \angle \mathrm{E}={50}^{°}$. Show that $\mathrm{AB}<\mathrm{BC}$.

$\mathrm{AD}$ bisects $\angle \mathrm{BAC}$ of $△\mathrm{ABC}$. Prove that $\mathrm{AB}>\mathrm{BD}$.

In a quadrilateral $ABCD$, show that $(AB+BC+CD+DA)>(AC+BD).$

In the quadrilateral $\mathrm{PQRS}, \mathrm{PQ}$ is the smallest side and $\mathrm{SR}$ is the longest side. Prove that $\angle \mathrm{P}>\angle \mathrm{R} \text{and} \angle \mathrm{Q}>\angle \mathrm{S}$.