Question

Given: ​​ ​​△ABC, FC¯¯¯¯¯ ∥BA¯¯¯¯¯, and A F¯¯¯¯¯ bisects ∠BAC


Prove: ABBD=ACCD

Answer 1

i just did the test so for future ppl

Explanation:

Answer 2

See proof

Explanation:

Statement Reason

1.
`\triangle ABC, \overline{FC}\parallel \overline {BA}` and
`\overline{FA}` bisects
`\angle BAC` - Given

2.
`\angle BAD\cong \angle CAD` - Definition of angle bisector

3.
`\angle BAD\cong \angle CFD` - Alternate interior angles theorem

4.
`\angle CFD \cong \angle CAD` - Substitution property

5.
`\bf{\angle ADB\cong \angle CDF}` - Vertical angles are congruent

6.
`\triangle ADB\sim \triangle FDC` - AA Similarity postulate

7.
`(AB)/(BD)=(FC)/(CD)` - Definition of similar triangles

8.
`AC=FC` - Converse of base angles theorem

9.
`(AB)/(BD)=(AC)/(CD)` - Substitution property