Japan Math Olympiad | A Very Nice Geometry Challenge 

It can be easily solved by using sine law in triangle ADC AND ABD as AD/sinθ =BD/sin140-θ and AD/sin20=AC/sin140 => Sin20/sinθ =sin140/sin140-θ => 2sin20sin(140-θ)= 2sinθsin140 => Cos(120-θ) - cos(160-θ) = cos(140-θ ) - cos(140+θ) =>2cos(50)cos(10+θ)=2cos(30+θ)cos10 Hence θ= 30

To a point E above AC, construct ∆EAC with the base AC and Congruent to ∆ABD. Since EC = AD = DC, and ∠ECD =60°, ∆ECD is Equilateral and ∆ADE is Isosceles. Since ∠ADE = 180-40-60=80°, ∠DAE = θ+20=½(180-80)=50° Hence, θ=30°

The angle theta is 30°. Also I think that this conatruction gets easier just from observing that congruent linea result congruent angles and you use the exterior angle theorem. After that just postulate that corresponding lines d angles are congruent by default. Then use the equilateral equiangular triangle to justify and triple-check that simplified line of reasoning. Then you use the alpha substitution and gets you theta equaling 30°. Maybe I just practice what I just thought of eventually.

AD=a..AC=b...teorema dei seni a/sinθ=b/sin(140-θ)...a/sin20=b/sin140...divido,elimino a e b..risulta ctgθ=(2cos20-cos40)/sin40..θ=30

{20°A+20°C+90°D}= 130°ACD {130°ACD ➖ 180°}= 50 5^10 5^2^5 1^2^1 2^1 (ACD ➖ 2ACD+1)

TRY ABD=70 IT WORKS TOO I CAN PROVE IT