Given: △ABC, FC¯¯¯¯¯ ∥BA¯¯¯¯¯, and A F¯¯¯¯¯ bisects ∠BAC
Prove: ABBD=ACCD
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Answer:
See proof
Step-by-step explanation:
Statement Reason
1. [tex]\triangle ABC, \overline{FC}\parallel \overline {BA}[/tex] and [tex]\overline{FA}[/tex] bisects [tex]\angle BAC[/tex] - Given
2. [tex]\angle BAD\cong \angle CAD[/tex] - Definition of angle bisector
3. [tex]\angle BAD\cong \angle CFD[/tex] - Alternate interior angles theorem
4. [tex]\angle CFD \cong \angle CAD[/tex] - Substitution property
5. [tex]\bf{\angle ADB\cong \angle CDF}[/tex] - Vertical angles are congruent
6. [tex]\triangle ADB\sim \triangle FDC[/tex] - AA Similarity postulate
7. [tex]\dfrac{AB}{BD}=\dfrac{FC}{CD}[/tex] - Definition of similar triangles
8. [tex]AC=FC[/tex] - Converse of base angles theorem
9. [tex]\dfrac{AB}{BD}=\dfrac{AC}{CD}[/tex] - Substitution property
Answer:
i just did the test so for future ppl
Step-by-step explanation: