Question

Given: ​​△ABC, F C¯¯¯¯¯ ∥B A¯¯¯¯¯, and A F¯¯¯¯¯ bisects ∠BAC

Prove: AB/BD=AC/CD

Answer 1

Answer:

had this exact question on my test! the correct answers are:

Alternate Interior Angles Theorem

<ADB~/=<CDF

Angle Angle Similarity Postulate

hope this helped!!

Answer 2

For given ΔABC,

$\rm \dfrac{AB}{BD}= \dfrac{AC}{CD}$

According to the given question

In the given ΔABC

$\rm \bar {FC} || \bar {BA}$

Also FA bisects the ∠BAC So we can conclude that

∠BAD = ∠DAC

∠ADB =  ∠ADC = 90°

So triangle ABD ≅ triangle ABC  ( According to angle angle similarity )

According to angle angle similarity if two angles of two triangles are congruent then triangles are said to be similar.

hence we can write the following

From Δ ABD and ΔABC

we can write

$\rm \dfrac{AB}{AC} = \dfrac{BD}{CD} \\\\ \\\\\dfrac{AB}{BD } = \dfrac{AC}{CD}$

For more information please refer to the link given below

https://brainly.com/question/10270676