Draw two triangles $ABC$ and $DBC$ on the same base and between the same parallels as shown in the figure with $P$ as the point of intersection of $AC$ and $BD$. Draw $CE\parallel BA$ and $BF\parallel CD$ such that $E$ and $F$ lie on line $AD$.

Can you show $ar(△PAB)=ar(△PDC)$

In $△ PQR, \angle PQR= 90°$and$\overline{QS} \perp \overline{PR }$ .The true relation among the following:

$$\begin{gathered} A) \overline{Q{S}^2} = \overline{PS} \times \overline{SR } \\ B)\overline{ P{Q}^2} =\overline{ RS} \times \overline{PR } \\ C) \overline{Q{R}^2 }= \overline{PS} \times \overline{ SR } \\ D) \overline{PQ} \times \overline{QR} =\overline{QS} \times \overline{PR} \end{gathered}$$

In $△ABC, AD$ is the median and $G$ is a point on $AD$ such that $AG:GD=2:1$. Then area $(△BDG):$area $(△ABC)$ is equal to:

In the figure, $BC=CD=DE$ and$P$  is mid-point of $CD$. The area of $△APC$ is 

If the bases of two triangles are situated on the same line and the other vertex of the two triangles are common, then the ratio of the areas of two triangles are _____ to the ratio of their bases.

Construct $△PQR$, in which $\angle Q=70°, \angle R=80°$ and $PQ+QR+PR=9.5 \mathrm{cm}$.

Construct $△XYZ$, such that $XY+XZ=10.3 \mathrm{cm}$, $YZ=4.9 \mathrm{cm},\angle XYZ=45°$.

Construct $△XYZ$, in which $YZ=6 \mathrm{cm}$, $XY+YZ=9 \mathrm{cm}$ and $\angle Y=50°$