Answer:

x = 20°

Step-by-step explanation:

The is problem is over 65+ years OLD and 6 ways to solve as of 2016. The diagram is actually flipped from the original. The following is one of the proofs based on its flipped version. Where A was B, B was C, C was A, D was E and E was D.

1. DRAW EF parallel to BC. Then angle DFE = 80º because of equal corresponding angles of parallel lines.

2. Drop a perpendicular line to BC from A, hitting BC at G. Because of congruent triangles ABG and AGC, angle BAG = angle CAG = 10º.

3. Now draw line FC, calling H the point where line FC intersects line BE. Line AG passes through point H, because of symmetry.

4. Angle BHC = 60º since the other angles of the triangle BHC are both 60º.

5. BE = FC (because of corresponding sides of congruent triangles FBC and EBC). BH = HC (call BH b) because triangle BHC is isosceles. So by subtraction, FH = HE.

6. Since angle FHE = 60º (vertical angle to BHC), and because FH =  HE from 5, angle FHE = angle HEF = 60º, so triangle FHE is equilateral. Thus, FE = FH = HE. Call each of those sides a.

7. Now AF = AE (because AB = AC and FB = EC, by subtraction AF = AE).

8. Because triangle AEB is isosceles, AE = BE = b + a. Thus, AF = BE = b + a (since AF = AE from 7).

9. BE = FC (congruent triangles BEC and BFC), so AF = FC, since AF = BE from 8.

10. Now watch this: Triangle AFH is congruent to triangle CFD because AF = FC; angle AFH = 140º = angle CFD;  angle DCF = 10º = angle FAH. Thus, corresponding sides of the congruent triangles AFH and triangle CFD are equal, so  FH = FD. But FH = FE from 6, so FE = FD.

11. Since FE = FD, angle FDE = angle FED and since angle DFE = 80º from (1), angle FDE = 50º = angle FED.

12. But angle FDC = 30º, so by subtraction, angle EDC = 20º!