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# Help Appreciated

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4 so I'm looking for lengths AE and CE

so I'm trying to figure out angle CAE and I determined that 180-DAB+CAD=CAE (DAB = CAD)

I also tried proving that angle CAE was 2*angle B by flipping and rotating the triangle to make a rectangle and then adding angles  ACD and B,

then I assumed because  CAe was on the same line and ACD + B made a 90 degree angle that CAE = ACD + B

The sidelengths of EA and CE are not equal

How should I continue

(if you have trouble deciphering my jumbled brain let me know, I'm not sure how to make a diagram for what I'm saying)

Feb 2, 2023
edited by YourAverageDummy  Feb 2, 2023

#1
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Let's see what i can do..

So you need lengths AE and CE

Apparently Angle ADC is not a right angle?

Feb 2, 2023
#2
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First, we know that $$DB =DC = \sqrt{9^2 - 3^2} = 6\sqrt{2}$$, which means $$CB = 12 \sqrt 2$$

Also, note that $$\triangle CEB \sim \triangle ADB$$ by AAA. This is because both triangles have a right angle and share $$\angle B$$, which means that all 3 traingles have the same angles, and thus, are similar.

Now, by similiar triangles, we have $${AB \over CB} = {AD \over CE}$$. Substituting what we know, we have $${9 \over 12 \sqrt 2} = {3 \over CE}$$, meaning $$CE = \color{brown}\boxed{4 \sqrt2}$$.

To solve for AE, we use the Pythagorean Theorem on $$\triangle AEC$$, and find that $$AE = \sqrt{9^2 - {4\sqrt2}^2} = \color{brown}\boxed{7}$$

Feb 2, 2023
#4
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Ohhhh so triangle  CEB is similar to the congruent right triangles in the middle

That makes sense thanks!

YourAverageDummy  Feb 2, 2023
#3
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Now that I think of it, here's another way of reaching the same answer.

We already know that $$CB = 12\sqrt2$$ (look at the other solution to see how)

Now, let $$CE = a$$ and $$AE = b$$. We have $$a^2 + b ^2 = 81$$, and $$a^2 + (b+9)^2 = ({12 \sqrt2})^2 = 288$$

Subtracting the two equations gives us $$18b + 81 = 207$$, meaning $$b = 7$$

Plugging this into the first equation also gives us $$a = 4 \sqrt 2$$

Feb 2, 2023