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Let C be a point not on line AE and D a point on line AE such that CD ⊥ AE. Meanwhile, B is a point on line CE such that AB ⊥ CE. If AB = 4, CD = 8, and AE = 5, then what is the length of CE? 

 

- Daisy

 Aug 8, 2018
edited by dierdurst  Aug 22, 2018
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\(\because \angle E = \angle E, \angle ABE = \angle CDE = 90º\therefore \text {By AA Similarity}, \triangle EBA \sim \triangle EDC.\\ \because \triangle EBA \sim \triangle EDC \therefore \frac{AB}{DC}=\frac{AE}{CE}.\\ \text{Filling in the numbers, we get}: \frac{4}{8}=\frac{5}{CE}.\\ \text{Solving for CE}: 4CE=5\cdot8\Rightarrow CE=10\)

 

The length of CE is 10. 

 

I hope this helped,

 

Gavin. 

 Aug 8, 2018

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