Question 1171816
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We're given that AB = AC, so triangle ABC is isosceles. This means the base angles B and C are congruent, as they are the base angles. Let y = angle B = angle C.
We're also given angle A is 60 degrees.
So,
A+B+C = 180
60+y+y = 180
2y+60 = 180
2y = 180-60
2y = 120
y = 120/2
y = 60
Each angle of triangle ABC is 60 degrees, so triangle ABC is equilateral.
This tells us that AB = BC = AC.


Draw a line from E to C
<img width="50%" src = "https://i.imgur.com/yyhKTzN.png">
We have similar triangles ADE and EDC
This means we can form the proportion
AD/DE = DE/DC
and that can be rearranged into
DE = sqrt(AD*DC)
This shows DE is the geometric mean of AD and DC


Let x = AD
Since AD is 1/4 the length of AC, we know that AC is 4 times longer compared to AD, so AC = 4*AD = 4x
This leads to 
AD+DC = AC
x+DC = 4x
DC = 4x-x
DC = 3x


Applying the geometric mean formula gets us
DE = sqrt(AD*DC)
DE = sqrt(x*3x)
DE = sqrt(3x^2)
This simplifies to x*sqrt(3), but we won't need to do this


With AD = x and DE = sqrt(3x^2), we can find AE through the pythagorean theorem
(AD)^2 + (DE)^2 = (AE)^2
(x)^2 + (sqrt(3x^2))^2 = (AE)^2
x^2+3x^2 = (AE)^2
4x^2 = (AE)^2
(AE)^2 = 4x^2
AE = sqrt(4x^2)
AE = sqrt((2x)^2)
AE = 2x


Now we can say
AE+EB = AB
2x+EB = 4x
EB = 4x-2x
EB = 2x
We know that AB = 4x because triangle ABC is equilateral.


The last step is to solve the proportion below
AE:EB = k:5
AE/EB = k/5
(2x)/(2x) = k/5
1 = k/5
k/5 = 1
k = 1*5
k = 5


So AE:EB = k:5 turns into AE:EB = 5:5 and we could reduce that ratio to 1:1
AE:EB = 1:1 indicates AE and EB are the same length.


Answer: A) 5
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