given that ab/de = bc/ef = 1/2, complete the statements to show that △abc ~ △def by the sas similarity theorem.
1st box: congruent, parallel, or perpendicular
2nd box: a and d, b and e, or c and f
3rd box: congruent, supplementary, or vertical

1st box ⇒ perpendicular

2nd box ⇒ B and E

3rd box ⇒ congruent

Step-by-step 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

1. Perpendicular

2. B and E

3. Congruent

Step-by-step explanation:

Horizontal and vertical lines are always perpendicular. So, the answer for the first gap is PERPENDICULAR.

Line BC is horizontal line, line AB is vertical line, so lines BC and AB are perpendicular and, therefore, angle B is right angle. Line EF is horizontal line, line ED is vertical line, so lines EF and ED are perpendicular and, therefore, angle E is right angle. So, answer for the second option is angles B and E.

Every two right angles are congruent.

Having two pairs of proportional sides and a pair of congruent adjacent angles, you can apply SAS similarity Postulate.

mk./km: lm

step-by-step explanation:

,mlmlm

; mk

lk

1/800 is the right answer