Let BC=a, AC=b, AB=c, BD=l, angle(BAC)=A, angle(ABC)=B, angle(ACB)=C. Then b=c and B=C, B=C=80. a=BC=2AB\cos(B)=2b\cos(80). AB^2=AD^2+BD^2-2AD.BD\cos(ADB)=AD^2+BD^2+2AB.BD\cos(CDB), where

BD^2=AB^2+AD^2-2AB.AD\cos(20)=b^2+a^2-2b.a\cos(20)=b^2+4b^2\cos^2(80)-4b^2\cos(80)\cos(20). Hence 2AB.BD\cos(CDB)=(AB^2-AD^2-BD^2)/2AB.BD or \cos(CDB)=[b^2-a^2-b^2-4b^2\cos^2(80)-4b^2\cos(80)\cos(20)]/{2b^2\sqrt[1+4\cos^2(80)-4\cos(80)\cos(20)]}=[b^2-4b^2\cos^2(80)-b^2-4b^2\cos^2(80)-4b^2\cos(80)\cos(20)]/{2b^2\sqrt[1+4\cos^2(80)-4\cos(80)\cos(20)]}=[-8\cos^2(80)-4\cos(80)\cos(20)]/{2\sqrt[1+4\cos^2(80)-4\cos(80)\cos(20)]}. Thus, \cos(CDB)=[-4\cos^2(80)-2\cos(80)\cos(20)]/{\sqrt[1+4\cos^2(80)-4\cos(80)\cos(20)]}, from where CDB can be obtained by converting the angles to radian and applying the inverse \arcos to both of the sides.

... construct an equilateral triangle BCE with side BC where point E is located inside the triangle ABC ... discover 2 identical triangles ... and, get the answer = 30 degrees = pi / 6 rad

... in fact, some purely geometric questions are ... 50+ (or more) years old ... in the eastern world ... so, good-long time for sharing it ...

Sándor Zoltán Németh Thank you for your interest in solving the problem. I am passionate about geometry and I propose to find a GEOMETRIC solution to this problem; you need to find the exact angle

another way ... construct an equilateral triangle ADE with side AD where point E is located outside the triangle ABC (on the right side of line AC) ... find 2 congruent triangles:

ABE and BAC

because AB is a common side, AE = BC, and the angle between these corresponding sides is 80 degrees, ... therefore BE=AC

next, triangle ADB is congruent to to triangle EDB

because both triangles have 3 equal sides,

thus, angle ABD = angle EBD = a half of 20 ... etc.

Liudmyla Hetmanenko, I imagined there must be an elegant geometric solution. If my calculations are correct, then combining the geometric solution and the analytic one, it follows a trigonometric identity. The solution of Steftcho P. Dokov is really nice. I really loved classical geometry as a kid and I shouldn't have jumped to a "brute-force" solution.

I have another solution. It is not so geometric, but it is relatively simple:

Let alpha=angle(CDB), beta=angle(ABD), gamma=angle(ADB), AD=BC=a, AB=AC=b.

I acknowledge that this may be above the level of school children, but perhaps there is a more elementary way to solve equation (3), an interesting trigonometric equation in itself.

Sorry, I was amazed by the beauty of this question, and I could not help of thinking about a trigonometric solution. Thanks for sharing this question Liudmyla Hetmanenko

Step 5. for the trigonometric solution above can be solved without derivatives as follows:

f(alpha)=sin(alpha-20)/sin(alpha)=[sin(alpha)cos(20)-cos(alpha)sin(20)]/sin(alpha)=cos(20)-cotan(alpha)sin(20). Hence, f is increasing and the equation (3): f(alpha)=2sin(10) can have at most one solution. But alpha=30 is a solution of (3). Thus, alpha is the unique solution of (3). Therefore, angle(CDB)=30.

Just for the sake of precision: The derivative formula in my original trigonometric solution is not correct. It either needs to be transformed in radians (20 replaced by pi/9), or multiplied by pi/180 . However, as shown above the derivative is not needed.