Question:
In the given figure, if ∠ADE = ∠ABC, then CE =
(a) 2
(b) 5
(c) 9/2
(d) 3
Solution:
Given: $\angle \mathrm{ADE}=\angle \mathrm{ABC}$
To find: The value of CE
Since $\angle \mathrm{ADE}=\angle \mathrm{ABC}$
$\therefore \mathrm{DE} \| \mathrm{BC}$ (Two lines are parallel if the corresponding angles formed are equal)
According to basic proportionality theorem if a line is parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.
In ∆ABC, DE || BC
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
$\frac{2}{3}=\frac{3}{E C}$
$\mathrm{EC}=\frac{3 \times 3}{2}$
$\mathrm{EC}=\frac{9}{2}$
Hence we got the result $(c)$.