+0

# help plz

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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

### 2+0 Answers

#1
+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   Jan 5, 2021
#2
+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