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

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

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

Ohhhh so triangle  CEB is similar to the congruent right triangles in the middle

That makes sense thanks!

YourAverageDummy  Feb 2, 2023

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

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