Question

HEY CAN ANYONE HELP ME!

Answer 1

Answer:

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

In ΔABC and ΔDBE

$DE || AC$   [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;

$\angle BDE \cong \angle BAC$  and

$\angle BEC \cong \angle BCA$

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

by Reflexive property of equality:

$\angle B \cong \angle B$  

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

$\triangle ABC \sim \triangle DBE$

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

then, we have;

$\frac{BD}{BA}$                  [By definition of similar triangles]

therefore, the missing statement and the reasons are

Statement                                                                   Reason

3.$\angle BDE \cong \angle BAC$      Corresponding angles theorem

and  $\angle BEC \cong \angle BCA$        

5. $\triangle ABC \sim \triangle DBE$   AAA similarity postulates    

6. BD over BA                                                    Definition of similar triangle

Answer 2

Here are the Statements and the Reason

3. <BDE congruent to <DAC    3. By corresponding angles
5. ΔABC is similar with ΔBDE    5. By AA proportionality
6. BD over BA is equal to BE over BC 6. Conclusion from 5