Triangle basic proportionally theorem

Question

In the given figure triangle ABC ~ triangle DEF . AP bisects angle CAB and DQ bisects angle FDE.

Prove that:

1) AP / DQ = AB / DE

2) triangle CAP ~ triangle FDQ

Answer 1

(i) \(\triangle ABC \sim \triangle DEF\)

\(\therefore \angle A = \angle D\quad{\text{(corresponding angles)}}\)

\(2\angle 1 = 2\angle 2\)

or, \(\angle 1 = \angle 2\)

Also,

\(\angle B = \angle E \quad{\text{(corresponding angles)}}\)

\(\triangle APB \sim \triangle DQE \)

or, \(\frac{AP}{DQ} = \frac{AB}{DE}\)

(ii) \(\triangle ABC \sim \triangle DEF\)

\(\therefore \angle A = \angle D\)

and \(\angle B = \angle E\)

or, \(2\angle 3 = 2 \angle 4\)

or, \(\angle 3 = \angle 4\)

\(\therefore \triangle CAP \sim \triangle FDQ \quad {\text{(By AA similarity)}}\)