BC = CD = DE = としてよい
∠A = 2x, ∠BEC = y とおく
△ABE が二等辺三角形だから
∠AEB = π/2 - x
△DCE が二等辺三角形だから
∠DCE = ∠DEC = π/2 - 2x, CE = 2sin(2x)
条件より
∠BCE = π - ∠A - ∠DCE = π/2
さらにBC = 1, CE = 2sin(2x)
∴ tan(y) = 1/(2sin(2x)) ... ①
条件より△CEFは直角三角形でCE = 2sin(2x), EF = 1/2 だから
cos(y+π/2-x) = 1/(4sin(2x)) ... ②
示すべき式は
0 = ∠A - ∠E = 5x - y - π
大先生に聞いたらz=0にはならんらしい

tan(y) = 1/(2sin(2x)), cos(y+π/2-x) = 1/(4sin(2x)) = 1/(4sin(2x)) , z = 5x - y - pi

https://ja.wolframalpha.com/input?i=tan%28y%29+%3D+1%2F%282sin%282x%29%29%2C+cos%28y%2B%CF%80%2F2-x%29+%3D+1%2F%284sin%282x%29%29+%3D+1%2F%284sin%282x%29%29+%2C+z+%3D+5x+-+y+-+pi