$\mathrm{ABCD}$ is a parallelogram. Prove that $∆\mathrm{ABD}\equiv ∆\mathrm{CDB}$.

$\mathrm{OD}$ is in the interior of $\angle \mathrm{AOB}$ . $\mathrm{P}$ is any point in $\mathrm{OD}, \mathrm{PC}, \mathrm{PD}$ are perpendicular on $\mathrm{OA},\mathrm{OB}$ respectively. If $\mathrm{PC}=\mathrm{PD}$, prove that$\angle \mathrm{POC}=\angle \mathrm{POD}$ .

$\mathrm{ABC}$ is an isosceles $∆$ . $\mathrm{BCDE}$ is a square. $\mathrm{AE}, \mathrm{AD}$ are joined. Prove that$∆\mathrm{ABE}\equiv ∆\mathrm{ACD}$ .

In the quadrilateral $\mathrm{ABCD}, \mathrm{AC}$ bisects $\angle \mathrm{A}$ and $\angle \mathrm{C}$. Prove that $\mathrm{AB}=\mathrm{AD}$ and $\mathrm{BC}=\mathrm{DC}$.

The side $\mathrm{AB}$ of $∆\mathrm{ABC}$ is produced on either side to $\mathrm{D}, \mathrm{E}$. Given that$\angle \mathrm{CBE}=148°$ and $\angle \mathrm{CAD}= 85°$, show that $\mathrm{BC}>\mathrm{AC}$ .

In$∆\mathrm{ABC}, \mathrm{AC} = 7 \mathrm{cm}$ and $\mathrm{BC}=8 \mathrm{cm}$. Show that $\mathrm{AB}<15 \mathrm{cm}$ and $\mathrm{AB}>1 \mathrm{cm}$ .

In $∆\mathrm{ABC}$, side $\mathrm{BA}$ is produced to $\mathrm{D}$and $\mathrm{CD}$ is joined. Given that $\mathrm{BA}= 5 \mathrm{cm}, \mathrm{AD}= 4 \mathrm{cm}$ and $\mathrm{BC}= 3.8 \mathrm{cm}$ show that$\mathrm{AC}<8.8 \mathrm{cm}$ and $\mathrm{CD}<12.8 \mathrm{cm}$.