SOLUTION: Given ΔABC, Segment AD bisects ∠BAC, Segment AE congruent to segment ED. I need help proving: AE/AC=BD/BC.
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Question 630966: Given ΔABC, Segment AD bisects ∠BAC, Segment AE congruent to segment ED. I need help proving: AE/AC=BD/BC.
If anyone can please explain and help me solve this. I would greatly appreciate it! I cannot figure this out, even after looking through my book or MML (My Math Lab)
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
AE congruent to ED
Therefore Angle EAD congruent to angle EDA
But Angle BAD congruent to EAD (AD is internal bisector)
Therefore angle BAD congruent to angle EDA.
But they are alternate angles .
Therefore Seg DE is parallel to Seg AB.
In triangle ABC AE/ EC = BD / DC [ Basic proportionality theorem)
EC/AE = DC/BD ( invertendo)
(EC+AE)/AE = (DC+BD)/BD ( componendo)
EC + AE = AC [ A-E-C]
DC+BD= BC [ B-D-C]
Therefore AC/AE = BC/BD
Therefore AE/AC = BD/BC ( invertendo)
Hence proved.
m.ananth@hotmail.ca
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