In the figure, given below, the medians $\mathrm{BD}$ and $\mathrm{CE}$ of a triangle $\mathrm{ABC}$ meet at $\mathrm{G}.$ Prove that : $∆\mathrm{EGD}~∆\mathrm{CGB}$

 

In the figure, given below, the medians $\mathrm{BD}$ and $\mathrm{CE}$ of a triangle $\mathrm{ABC}$ meet at $\mathrm{G}.$ Prove that :

$\mathrm{BG}=2 \mathrm{GD}$ from $\left(\mathrm{i}\right)$ above.

In the following figure, point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio $3:5.$ Find: $\frac{\mathrm{AE}}{\mathrm{EC}}$

In the following figure, point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio $3:5.$ Find :

$\frac{\mathrm{AD}}{\mathrm{AB}}$

In the following figure, point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio $3:5.$ Find:

$\frac{\mathrm{AE}}{\mathrm{AC}}$

In the following figure, point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio $3:5.$ Find:

$\mathrm{DE}=2.4\mathrm{cm},$ find the length of $\mathrm{BC}.$

In the following figure, point $\mathrm{D}$ divides $\mathrm{AB}$ in the ratio $3:5.$ Find:

$\mathrm{BC}=4.8\mathrm{cm},$ find the length of $\mathrm{DE}.$

In the given figure, $\mathrm{PQ}\parallel \mathrm{AB};$ $\mathrm{CQ}=4.8\mathrm{cm} \mathrm{Qb}=3.6\mathrm{cm}$ and Find :

$\frac{\mathrm{CP}}{\mathrm{PA}}$

In the given figure, $\mathrm{PQ}\mathit{\parallel }\mathrm{AB}$; $\mathrm{CQ}=4.8\mathrm{cm} \mathrm{QB}=3.6\mathrm{cm}$ and $\mathrm{AB}=6.3\mathrm{cm}$ Find :

$\mathrm{PQ}$