Challenging Geometry Reasoning

In $\triangle BDE$ , $\angle BDE = 45^{\circ}$ since $\overline{BD }$ is the angle bisecting line.

Using sine rule,

$\frac{DE}{\sin 22.5} = \frac{BE}{\sin 45} \tag{1}$

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In $\triangle BEC$ , $\angle BCE = 90^{\circ}$ as it is the squares edge. (square property).

Using the sine rule here we obtain,

$\frac{EC}{\sin 22.5} = BE \tag{2}$

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Letting $BE$ subject in $(1)$ and $(2)$ we attain,

$BE = \frac{ \sin 45 DE}{\sin 22.5} \tag{1}$ $BE = \frac{EC}{\sin 22.5} \tag{2}$

Thus BE = BE and we attain,

$\frac{ \sin 45 DE}{\sin 22.5} = \frac{EC}{\sin 22.5}$

Multiplying both sides by $\sin 22.5$ we obtain, ( note that $\sin 45$ = $\frac{\sqrt{2}}{2}$ ).

$\boxed{\overline{EC} = \frac{\sqrt{2}}{2} \overline{DE}}$

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Find point F on line BD such that it forms a right triangle DFE. The rest follows, as FE = EC.

tan 22.5 = sqrt(2)-1 Thus, tan22.5= EC/BC Thus, BC = EC/(sqrt(2)-1) Also, tan 45 =1 and thus, DC = BC = DE + EC Thus, DE+EC = EC/(sqrt(2)-1) Thus sqrt(2)DE - DE + sqrt(2)EC-EC=EC. Thus, sqrt(2)EC(sqrt(2)-1)=DE(sqrt(2)-1). Thus, EC=1/sqrt(2) DE = sqrt(2)/2 DE