In triangle DEllBC, AD=1cm and BD

It is given that AD = 1 cm,
BD = 2 cm and {tex}D E \| B C{/tex}
In  {tex}\triangle{/tex}ADE and {tex}\triangle {/tex}ABC
{tex}\angle A D E = \angle A B C{/tex} (Corresponding angles)
{tex}\angle A = \angle A{/tex} [Common]
Therefore, by A.A. similar condition
{tex}\triangle \mathrm { ADE } \sim \triangle \mathrm { ABC }{/tex}
Ratio of areas of similar triangles is equal to the square of the ratio of the corresponding sides.
{tex}\therefore \quad \frac { \operatorname { ar } ( \triangle \mathrm { ABC } ) } { \operatorname { ar } ( \triangle \mathrm { ADE } ) } = \frac { \mathrm { AB } ^ { 2 } } { \mathrm { AD } ^ { 2 } }{/tex}
{tex}\Rightarrow \quad \frac { \operatorname { ar } ( \triangle A B C ) } { \operatorname { ar } ( \triangle A D E ) } = \left( \frac { 3 } { 1 } \right) ^ { 2 }{/tex}
{tex}\Rightarrow \quad \frac { \operatorname { arc } \triangle A B C ) } { \operatorname { ar } ( \triangle A D E ) }{/tex}
{tex}= \frac { 9 } { 1 }{/tex}