Answer:
[tex]\text{Solution:}\\1.\ \text{In triangles AEF and ABC}\\\text{}\ \ \ \text{i. }\angle \text{A}=\angle \text{A (A)}\ \ \ [\text{Common angle in both triangles.}]\\\text{}\ \ \ \text{ii. }\angle \text{AEF}=\angle \text{ABC (A)}\ \ \ [\text{EF}\parallel\text{BC, corresponding angles.}]\\\text{}\ \ \ \text{iii. }\triangle \text{AEF}\sim\triangle\text{ABC}\ \ \ [\text{By A.A. postulate}][/tex]
[tex]2.\ \frac{\text{AB}}{\text{EA}}=\frac{\text{AC}}{\text{FA}}\ \ \ [\text{Corresponding sides of similar triangles are proportional.}]\\\\\text{or, }\frac{\text{AE+EB}}{20}=\frac{\text{FA+FC}}{\text{FA}}[/tex]
[tex]\text{or, }\frac{\text{20+4}}{\text{4}}=\frac{\text{FA+3}}{\text{FA}}\\\\\text{or, 6}=(\text{FA+3}/\text{FA})\\\text{or, 6FA = FA + 3}\\\text{or, 5FA = 3}\\\text{or, FA = 3/5}\\\therefore\ \text{FA = 0.6}[/tex]
