Question

the bisector of exterior angle CAF of triangle ABC intersects the side BC produced at D show that BA /AC BD/DC

Answer 1

Answer:

Step-by-step explanation:

Given: The bisector of exterior angle CAF of triangle ABC intersects the side BC produced at D.

To prove:BA /AC BD/DC

Construction: Draw CE║AB which meets AB at E.

Proof: In ΔABC, CE║AD which is cut by AC,

∠DAC=∠ECA( Alternate angles)

Similarly, CE║AD which is cut by AB,

∠DAF=∠CEA( Corresponding angles)

and AC=AE( by isosceles triangle theorem)

Now, in ΔBAD, we have CE║DA

⇒$\frac{AE}{AB}=\frac{DC}{BD}$( by basic proportionality theorem)

But, we have proved above that AC=DC

⇒$\frac{BA}{AC}=\frac{BD}{DC}$

Hence proved.