Answer:
1st box ⇒ perpendicular
2nd box ⇒ B and E
3rd box ⇒ congruent
Explanation:
* Lets revise the case SAS of similarity
- SAS similarity : In two triangles, if two sets of corresponding sides
are proportional and the included angles are equal then the two
triangles are similar.
- Example : In triangle ABC and DEF, if m∠A = m∠D and
AB/DE = AC/DF then the two triangles are similar by SAS
* Lets solve the problem
- In Δ ABC and Δ DEF
∵ AB/DE = BC/EF = 1/2
- That means two sets of corresponding sides are proportion
∵ AB is a vertical side and BC is a horizontal side
∵ DE is a vertical side and EF is a horizontal side
∵ Horizontal and vertical lines are perpendicular
∴ AB ⊥ BC and DE ⊥ EF
- So angles B and E are right angles by definition of perpendicular
lines
∵ All right angles are congruent
∴ m∠B = m∠D
∵ The two triangles have two sets of corresponding sides are
proportion and the included angles are equal then the two
triangles are similar
∴ △ABC ~ △DEF by the SAS similarity theorem