Which of the following completes the proof? triangles abc and edc are formed from segments bd and ac, in which point c is between points b and d and point e is between points a and c. given: segment ac is perpendicular to segment bd prove: δacb ~ δecd reflect δecd over segment ac. this establishes that then, this establishes that ∠e'd'c' ≅ ∠abc. therefore, δacb ~ δecd by the aa similarity postulate. ∠abc ≅ ∠e'd'c'; translate point e' to point a ∠acb ≅ ∠e'c'd'; translate point e' to point b ∠acb ≅ ∠e'c'd'; translate point d' to point b ∠abc ≅ ∠e'd'c'; translate point d' to point a

∠ACB ≅ ∠E'C'D'; translate point D' to point B

Step-by-step explanation:

That is just my best guess.

∠ACB ≅ ∠E'C'D'. translate point E' to point A

Step-by-step explanation:

the answer for this question is : 360°

step-by-step explanation:

we have been given a triangle   we need to tell which one of these is a triangle below to map onto itself.

360° rotation is the complete rotation which eventually leads to the same figure which is given.

the correct will be, that is 360° that will cause the triangle below to map onto itself.

1. 16.64

a^2+b^2=c^2

pr^2=pq^2+qr^2

pr^2=14^2+ 9^2

pr^2=196+81

pr= sqr root of 277

16.64

2. 11.31

c^2-a^2=b^2

18^2- 14^2=b^2

b^2=324-196

b=sqr root of 126

b=11.31

hope it ! (btw im in connections academy to, good luck! )