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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=15, and EB=30.

 Jan 5, 2021
 #1
avatar+114096 
+1

                   B

                   l

                   l

                   l

                   l

           30    l

                   l

                   l

        E         l

             15  l

                   P

                   l      10

A                D              C

 

BP =sqrt (30^2 +15^2)   = sqrt (1125)= 15sqrt (5)

 

sin BPE =  30 / (15 sqrt (5))  = 2/sqrt (5)  = sin CPD

 

sin PCD  =  sqrt ( 1 - (2/sqrt (5))^2)  =  sqrt ( 1 -4/5)   =  1/sqrt (5)

 

PD / sin PCD =  PC /sin 90

 

PD =  10 / sqrt (5) =   2sqrt (5)

 

BD =  17sqrt (5)

 

angle  BPE =  angle BAD

 

BD / sin BAD = BA / sin  BDA

 

17sqrt (5) /( 2 /sqrt (5))  =  (30 +AE) / sin 90

 

(17 * 5)/2  = 30 + AE

 

85/2   =  30 + AE

 

42.5  = 30  + AE

 

AE =  42.5  - 30   =  12.5

 

 

cool cool cool

 Jan 5, 2021
 #2
avatar+824 
+2

Triangle ABC is acute. Point D lies on AC so that line BD is perpendicular 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=15, and EB=30.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

ΔACE ~ ΔBPE

 

AE / CE = PE / EB               AE = (25 * 15) / 30

 

AE = 12.5

 Jan 5, 2021
edited by jugoslav  Jan 5, 2021

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