Çözüldü Angle in Right Triangle - Pythagorean Theorem - Quadratic Equation - Trigonometry

A proposed SAT Math Grid-in Question:

ABC is a right triangle. Angle C is 90 degrees. There is a point D on the side between A and C. If |AD| = |BD| and |AB| = |AC| + |CD|, what is angle A in radians?

α = ∡A
|AD| = p
|CD| = q

Synthetic Solution:
Pick a point E on [AC[ so that |DE| = |CE|, ∡ABD = ∡α, ∡BDE = 2α = ∡CEB = ∡AEB, ∡DBC = 90° - 2α = ∡CBE
|AB| = p + q + q = p + 2q = |AE|, so triangle ΔABE is isosceles, and ∡ABE = ∡AEB which means α + 2(90° - 2α) = 2α
180° - 4α = α
180° = 5α
π / 5 Radians = α.

Algebraic Trigonometry Solution:
Applying Pythagorean Theorem: p^2 - q^2 = (p + 2q)^2 - (p + q)^2 ⇒ 4(q / p)^2 + 2(q / p) - 1 = 0
q / p = (√5 - 1) / 4....(I)
In the right triangle BCD, cos(2α) = q / p....(II)
From the equations (I) and (II), 2·[ cos(α) ]^2 - 1 = (√5 - 1) / 4
cos(α) = [ (√5 + 3) / 8 ]^0.5....(III)
(√5 + 3) / 8 = [ (√5 + x) / y ]^2 and solving on the set of non-negative integers yields x = 1, y = 4, so;
cos(α) = (√5 + 1) / 4 ====> α = The famous 36° ====> π / 5 Radians.

Notes:
1. On the calculator-allowed section of the SAT exam, from the equation (III) directly α = arccos{ [ (√5 + 3) / 8 ]^0.5 } = 36°
https://www.wolframalpha.com/input?i=arccos{ [ (√5 + 3) / 8 ]^0.5 } =
2. Original Turkish version of the problem is at https://www.facebook.com/photo/?fbid=208298336009653&set=g.157000044359638

The solution by the AI at www.mathful.com is attached.