geometry problem1

Let $|\overline {CD}|=x, |\overline {EC}|=y,|\overline {BE}|=z,\angle {BDC}=α$

Then $x=(6+y)\cos 40\degree$

$\dfrac {z}{\sin 100\degree}=\dfrac {4}{\sin (40\degree+α)}=\dfrac {y}{\sin (40\degree-α)}$

$\dfrac {4}{\sin α}=\dfrac {x}{\sin (40\degree-α)}$

From these we get

$\tan α=\dfrac {4-y}{4+y}\tan 40\degree=\dfrac {4}{10+y}\tan 40\degree$

$\implies y=2,x=8\cos 40\degree,\tan α=\dfrac 13\tan 40\degree$

So, $z=\dfrac {\sin 100\degree\sqrt {9+\tan^2 40\degree}}{\sin 40\degree}$

Hence $2z^2-x^2=\dfrac {2\sin^2 100\degree(9+\tan^2 40\degree)}{\sin^2 40\degree}-64\cos^2 40\degree=\boxed 8$ .