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# Kind of a harder one. Full picture

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I really just dont know where to start on this one to be honest. I thought about making a triangle BEA or CBG but not sure how that would work. Thansk for the help.

Feb 6, 2020

#1
+2857
+2

\

Aha! CBD ~ CGA!

Both are 30-60-90.

1. Set CG = $$x$$, and set CB = $$y$$

2. CD is $$x+1$$

3. Based on 30-60-90 properties, $$\frac{x+1}{2}=y$$

4. Write a similarity proportion

$$\frac{y}{x+1}=\frac{x}{y+13}$$

So now we have two equations...

Have fun solving!

Feb 6, 2020
#3
+2

I actually got this far after 2 hours but I can't figure out EA. Here's what I have thus far... GA = 6.5 + root3/3, CD = 5.08611, the intersection of the auxilliary line you drew and line FGA I called H. HA is 6.5.

Guest Feb 6, 2020
#2
+25
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Here is my solution :

https://imgur.com/a/hhC4nVl

I just used some basic geometry theories like Equilateral Triangle, Pythago, etc.

Hope it can help :)

Feb 6, 2020
#4
+822
+4

Triangles  BCD, ACG, and GDN are similar;  all of them are  30-60-90

Let's denote the intersection point of AF and BD with a leter N

BN = 13 * tan(30°) = 7.505

DN = 1 / cos (30°)  = 1.155

BD = BN + DN = 8.66

CD = BD / cos (30°) = 10

Circle radius r = CD / 2 = 5

EG = sqrt [ r² - ( r - 1 )² ] = 3

CG = CD - GD = 9

AG = CG / tan (30°)  = 15.588

EA = AG - EG = 12.588

Feb 6, 2020
edited by Dragan  Feb 6, 2020
edited by Dragan  Feb 6, 2020
edited by Dragan  Feb 6, 2020
#6
+109806
+4

Thanks guys,

That is really impressive.

I followed your logic and then drew the pic.

Here it is.

I have coloured the 3 similar triangles that are used in the solution.

Feb 6, 2020