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.
+128470 B
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30 l
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15 l
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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
+1639 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.
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ΔACE ~ ΔBPE
AE / CE = PE / EB AE = (25 * 15) / 30
AE = 12.5
