Answer:
Yes, △ABC ∼ △FED by AA postulate.
Step-by-step explanation:
Given:
Two triangles ABC and FED.
m∠A = m∠B
m∠C = m∠A + 30°
m∠E = m∠F = [tex]x[/tex]
m∠D = [tex]2x-20[/tex]°.
Now, let m∠A = m∠B = [tex]y[/tex]
So, m∠C = m∠A + 30° = [tex]y+30[/tex]
Now, sum of all interior angles of a triangle is 180°. Therefore,
m∠A + m∠B + m∠C = 180
[tex]y+y+y+30=180\\3y=180-30\\3y=150\\y=\frac{150}{3}=50[/tex]
Therefore, m∠A = 50°, m∠B = 50° and m∠C = m∠A + 30° = 50 + 30 = 80°.
Now, consider triangle FED,
m∠D+ m∠E + m∠F = 180
[tex]2x-20+x+x=180\\4x=180+20\\4x=200\\x=\frac{200}{4}=50[/tex]
Therefore, m∠F = 50°
m∠E = 50° and
m∠D = [tex]2x-20=2(50)-20=100-20=80\°[/tex]
So, both the triangles have congruent corresponding angle measures.
m∠A = m∠F = 50°
m∠B = m∠E = 50°
m∠C = m∠D = 80°
Therefore, the two triangles are similar by AA postulate.