In the figure, $AB$ and $DE$ are perpendicular to $BC$.
- Prove that $\triangle ABC \sim \triangle DEC$.
- If $AB = 6 \; cm$, $DE = 4 \; cm$, $AC = 15 \; cm$, calculate $CD$.
- Find the ratio of $\;\;$ $\dfrac{\text{Area}\left(\triangle ABC\right)}{\text{Area}\left(\triangle DEC\right)}$
- Given: $\;\;$ $AB \perp BC$, $\;$ $DE \perp BC$
$\therefore \;$ $\angle ABC = \angle DEC = 90^\circ$ $\;\;\; \cdots \; (1)$
In triangles $ABC$ and $DEC$,
$\angle ACB = \angle DCE$ $\;$ [common angle] $\;\;\; \cdots \; (2)$
$\therefore \;$ From equations $(1)$ and $(2)$, $\;$ $\triangle ABC \sim \triangle DEC$ $\;$ [by Angle-Angle postulate] - $\because \;$ $\triangle ABC \sim \triangle DEC$
$\therefore \;$ $\dfrac{AB}{DE} = \dfrac{AC}{DC}$ $\;\;\;$ [corresponding sides of similar triangles are in proportion]
$\therefore \;$ $DC = \dfrac{AC \times DE}{AB} = \dfrac{15 \times 4}{6} = 10 \; cm$ - $\dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area}\left(\triangle DEC\right)} = \dfrac{AB^2}{DE^2}$
[areas of two similar triangles are proportional to the squares on their corresponding sides]
i.e. $\;$ $\dfrac{\text{Area} \left(\triangle ABC\right)}{\text{Area}\left(\triangle DEC\right)} = \dfrac{6^2}{4^2} = \dfrac{9}{4}$