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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=20, and EB=30.

 Oct 12, 2018
 #1
avatar+101872 
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We have that

 

Δ ADB  ~ Δ AEC      and

 

Δ ADB  ~ Δ PEB

 

So

 

ΔAEC  ~   Δ PEB

 

AE / EC  = PE / EB

 

AE / [CP + EP ]    = PE / EB

 

AE  / [ 10 + 20 ]  = 20 / 30

 

AE / 30  = 20 / 30  ⇒   AE   = 20

 

 

cool cool cool

 Oct 12, 2018

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