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Triangle ABC is acute. Point D lies on AC so that line BD is perperdicular to line AC, and point E lies on AB such that line CE is perpendicular to line AB . The intersection of segments line CE and line BD is P. Find the value of AE if CP=10, PE=25, and EB=30.

 Jan 4, 2021
 #1
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Δ ADB  ~ Δ AEC      

 

Δ ADB  ~ Δ PEB

 

ΔAEC  ~   Δ PEB

 

AE / EC  = PE / EB

 

AE / [CP + EP ]    = PE / EB

 

AE  / [ 10 + 20 ]  = 20 / 30

 

AE / 30  = 20 / 30  ⇒   AE   = 20

 

20!

 Jan 4, 2021
 #2
avatar+266 
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All right, this is getting ridiculous. Melody and I have told you many times to not copy others' answers. This time you copied CPhill's. We both know very well that plagarism is bad, so just put a link...

 

cheekycheekycheeky

Here is CPhill's answer: https://web2.0calc.com/questions/help-fast-please_4

DewdropDancer  Jan 4, 2021

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