Hey can anyone me!
in δabc shown below, segment de is parallel to segment ac:
triangles abc and dbe where de is parallel to ac
the two-column proof with missing statements and reasons proves that if a line parallel to one side of a triangle also intersects the other two sides, the line divides the sides proportionally:
statement reason
1. line segment de is parallel to line segment ac 1. given
2. line segment ab is a transversal that intersects two parallel lines. 2. conclusion from statement 1.
3. 3.
4. ∠b ≅ ∠b 4. reflexive property of equality
5. 5.
6. bd over ba

Given: In triangle ABC and triangle DBE where DE is parallel to AC.

In ΔABC and ΔDBE

  [Given]

As we know, a line that cuts across two or more parallel lines.  In the given figure, the line AB is a transversal.

Line segment  AB is transversal that intersects two parallel lines.  [Conclusion from statement 1.]

Corresponding angles theorem: two parallel lines are cut by a transversal, then the  corresponding angles are congruent.

then;

 and

Reflexive property of equality states that if angles in geometric figures can be congruent to themselves.

by Reflexive property of equality:

By AAA (Angle Angle Angle) similarity postulates states that all three pairs of corresponding angles are the same then, the triangles are similar

therefore, by AAA similarity postulates theorem

Similar triangles are triangles with equal corresponding angles and proportionate side.

then, we have;

                 [By definition of similar triangles]

therefore, the missing statement and the reasons are

Statement                                                                   Reason

3.      Corresponding angles theorem

and          

5.   AAA similarity postulates    

6. BD over BA                                                    Definition of similar triangle

(0.2)+(-8)=0.3+-3+0.3m is the equation i guess? ?

step-by-step explanation: