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.
+87 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!
+240 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...
Here is CPhill's answer: https://web2.0calc.com/questions/help-fast-please_4
