Answer:
60°
Step-by-step explanation:
Extend AB beyond B, extend the segment with point D beyond D so it intersects BC at point E, and extend CD beyond D so that it intersects the extension of AB at point F (see attachment).
Now, we have two triangles: ΔCBF and ΔCED. Because DE || BF, then the two triangles are similar from equal angles.
ABF is a line, so ∠ABF = 180. We also know that ∠ABF = ∠ABC + ∠CBF = 130 + ∠CBF = 180. Then ∠CBF = 50°. Because the two triangles are similar and ∠CBF corresponds with ∠CED, then ∠CED = ∠CBF = 50°.
DE is also a line, so it is 180°. From similar reasoning as above, ∠CDE = 180 - 110 = 70°.
Since CDE is a triangle, its angles add up to 180°, so:
∠CED + ∠ECD + ∠CDE = 180
50 + ∠ECD + 70 = 180
∠ECD = ∠BCD = 60°
The answer is 60°.
