Question

the bisector of the exterior angle C A F of a triangle ABC, intersect the side BC produced at D.Show that BA/AC = BD/DC

Answer 1

Answer:

This result is proved with the help of side splitter theorem corollary.

Step-by-step explanation:

Given the bisector of the exterior angle C A F of a triangle ABC, intersect the side BC produced at D. we have to prove that $\frac{BA}{AC}=\frac{BD}{DC}$

In the figure, in order to prove required result let we do construction, draw a line parallel to AD intersect the line AB as shown.

implies ∠DAC = ∠ACE ( ∵ Alternate angles)

            ∠DAF=∠CEF  ( ∵ Corresponding angles)

Now, because ∠DAC = ∠DAF (Given)

∠ACE=∠CEF

Hence, ACE is an isosceles triangle ∴ AE=AC

Therefore, by side splitter theorem corollary, we have

$\frac{BA}{EA}=\frac{BD}{CD}$

∵ EA=AC

Hence,  $\frac{BA}{AC}=\frac{BD}{DC}$