Halp

can we assume that AE and BE are the same length? if so, DE should be 4cm.

Nope the triangle in the middle is not isosceles

Global rule 2 etc.

Solving for (where [math]\alpha = AED [/math]:

[math]DE = \frac{9\cos \alpha}{\sin \alpha} [/math]

Where [math]\alpha [/math] is such that [math] \sin \left(2\alpha - \frac{\pi}{4} \right) = \frac{5}{4\sqrt{2}} [/math]

Solve for alpha by putting that into Wolfram Alpha, then plug that into the equation for DE

You get DE around 6.6

I'll post general working out soon

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>>126878

part 1
if any of this goes over your head completely (i.e. not familiar at all) then you cannot do this question

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>>126886