Hello kastamonu,
using your labels.At C a semi-circle is drawn thru A and B meeting AE extended at D and AC extended at H. Note the following
arc AB =60,arc BD =2*40= 80, arc DH = 2*20 =40. Angle BCD = 80 (unknown angle)
Triangle ADC is isosceles.AC =CD (radii). Angle BEA is 80. DEC is 80 (vertical angles)
Triangle EDC is isosceles (base angles both 80- angles DEC and BCD)
DE=DC =AB as specified in given info
Hello,
I've been working on this problem with no success. Idea, could you be more explicit in your proof of similarity of the triangles?
Kastomonu, it's a language translation problem. Could you draw a diagram that illustrates what "Draw a symmetric of AE according to A" means?
AE/EC = AB/FC turned out to be a bit complicated using properties of 40, 60, 80 triangles, Law of cosines, etc
Here is a much easier method.
BG is the bisector of angle ABE, G on AC. angle BGE = 80 and triangle BGE is isosceles, so let b = BG = BE and a = AB = BC = CA = ED
GE/EC = b/a since BE is a bisector in triangle BCG
BF/FC = b/a since EF is parallel to CD
so GE/EC = BF/FC therefore EF is parallel to BG and angle ECD = angle FEC = angle BGE = 80