In the given figure, if $\angle ADE=\angle B$, show that $\Delta ADE\sim \Delta ABC$. If AD = 3.8 cm, AE = 3.6 cm, BE = 2.1 cm and BC = 4.2 cm, find DE.

In $\Delta ADEand\Delta ABC,$

$\angle A=\angle A\left(\text{Common angle}\right)$

$\angle ADE=\angle B\left(\text{Given}\right)$

So by AA similarity, triangles are similar.

So, by CPCT,

$\frac{AD}{AB}=\frac{DE}{BC}$

$\frac{3.8}{3.6+2.1}=\frac{DE}{4.2}$

$DE=\frac{3.8\times 4.2}{5.7}=2.8$ (2 marks)