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.

Guest Jan 4, 2021

#1**0 **

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!

JoLink Jan 4, 2021

#2**+1 **

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

DewdropDancer
Jan 4, 2021