The figure below shows segments ac and ef which intersect at point b. segment af is parallel to segment ec: which of these facts is used to prove that triangle abf is similar to triangle cbe? angle fab is equal to angle ceb because corresponding angles are congruent. angle abf is congruent to angle ceb because vertically opposite angles are congruent. angle afb is congruent to angle ceb because alternate interior angles are congruent. angle afb is congruent to angle ceb because supplementary angles are congruent. i chose b, but i'm not positive it is the answer. can anyone give some clarification?

 Angle AFB is congruent to angle CEB because alternate interior angles are congruent.

Step-by-step explanation:

Given: ,

And, ac and ef are intersecting each other at point b.

Prove: Triangle abf is similar to triangle cbe

Since,   (Reflexive)

,

⇒ ef is the common transversal of parallel lines fa and ec.

        (Because Alternative interior angles are congruent)

Thus, By AA similarity postulate,

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.

x-y=3

4x+y=-3

5x=0

x=0

then substitute into one equation to find y.

y=-4(0)-3

y=-3