Let BD=DC=a and draw a vertical line AE so E is left of B. Let AE=z and EB=y. Then (y+2a)/z = cot(30) = tan (60) = sqrt (3) and (y+a)/z= cot (45) = tan (45) = 1. Then a/z = tan (60) - tan (45) = sqrt (3) - 1, thus y/z = 1 - a/z = 2-sqrt (3) = tan (BAE) = tan (P) -> P=15. Now P+Theta = 90-45 = 45 --> Theta = 45-15 = 30.
I used Pyphagorean theorem for four triangles: one is buit by the perpendicular on the base line(the same as you did), two are from a perpendicular on AD, and the fourth is APB . I calculated PB and then defined the sin of theta. it is1/2.
At 4:37, Math Booster has found that y/b = 1/(2 - √3). y/b is the ratio of sides for ΔABP. We can recognize that 1/(2 - √3) is one way of expressing the ratio of sides, (long side)/(short side) in a 15°-75°-90° right triangle. So,
Method 1: As ∠BDA = 45°, ∠ADC = 180°-45° = 135°. As ∠BDA is an exterior angle to ∆DCA at D, ∠CAD = 45°-30° = 15°. Draw DE, where E is the point on CA where ∠ADE = 15°. By construction, ∆DEA is an isosceles triangle, so EA = DE and ∠DEA = 180°-(15°+15°) = 150°. As ∠DEA = 150°, ∠CED = 180°-150° = 30°. As ∠DCA = 30° as well, ∆EDC is an isosceles triangle and DC = DE. As ∠BDA = 45° and ∠ADE = 15°, ∠BDE = 60°. As BD = DE, then ∆DEB is an equilateral triangle, EB = BD = DE, and ∠DEB = ∠EBD = 60°. As BE = EA, ∆BEA is an isosceles triangle. As ∠CEB = 30°+60° = 90°, ∠BEA = 90°, and ∠ABE = ∠EAB = (180°-90°)/2 = 45°. As ∠CAD = 15°, ∠DAB = θ = 45°-15° = 30°. Method 2: Extend DB and drop a perpendicular from A to E, so that DE and BE are collinear. As ∠ECA = 30° and ∠AEC = 90° by construction, ∠CAE = 90°-30° = 60° and ∆AEC is a 30-60-90 special right triangle. If AE = a, then CA = 2a and EC = √3a. As ∠EDA = 45°, ∠DAE = 90°-45° = 45° as well, ∆AED is an isosceles right triangle, and AE = ED. As AE = a, DC = √3a-a = (√3-1)a. Additionally, DA = √(a²+a²) = √2a. As ∠BDA = 45°, ∠ABD = 135°- θ. By the law of sines: √2a/sin(135°- θ) = (√3-1)a/sin(θ) √2asin(θ) = (√3-1)asin(135°- θ) sin(135°- θ) = √2sin(θ)/(√3-1) sin(135°- θ) = sin(135°)cos(θ) - cos(135°)sin(θ) sin(135°- θ) = cos(θ)/√2 + sin(θ)/√2 sin(135°- θ) = (cos(θ)+sin(θ))/√2 (cos(θ)+sin(θ))/√2 = √2sin(θ)/(√3-1) 2sin(θ) = (cos(θ)+sin(θ))(√3-1) 2sin(θ) = (√3-1)cos(θ) + (√3-1)sin(θ) 2sin(θ) - (√3-1)sin(θ) = (√3-1)cos(θ) (3-√3)sin(θ) = (√3-1)cos(θ) √3(√3-1)sin(θ) = (√3-1)cos(θ) √3sin(θ) = cos(θ) tan(θ) = 1/√3 ==> θ = 30°
