Given:   `deltaabc`  with  `bar(de)`  ||  `bar(ac)`  prove:   `(ad)/(db) = (ce)/(eb)`statementreason1.  `bar(de)`  ||  `bar(ac)`given2.  `/_ cab~=/_edb`        `/_acb~=/_deb`if two parallel lines are cut by a transversal, the corresponding angles are congruent.3.  `deltaabc ~ deltadbe`aa criterion for similarity4.  corresponding sides of similar triangles are proportional.5.  ab = ad + db      cb = ce + ebsegment addition6.  `(ad + db) /(db) = (ce + eb) /(eb)`substitution property of equality7.  `(ad)/(db) + 1 = (ce)/(eb) + 1`division8.  `(ad)/(db) = (ce)/(eb)`subtraction property of equalitywhat is the missing statement in the proof?   scroll down to see the entire proof.

Given: ΔABC with DE║AC

To Prove:

Statements and Reasons

1. DE║AC (given)

2. ∡CAB = ∡EDB and ∡ACB = ∡DEB (If two parallel lines are cut by a transversal, the corresponding angles are congruent.
)

3. ΔABC ≈ ΔDBE (Angle-Angle Similarity theorem)

4. (Corresponding sides of similar triangles are proportional.)

5. AB = AD + DB
and CB = CE + EB
(segment addition postulate)

6. (Substitution Property of Equality
)

7.
(Division property)

8. (Subtraction Property of Equality
)

So missing statement in the proof was

answer:

given the linear inequality :

add 2y to both sides we get;

simplify:

add 12 to both sides of an equation;

simplify:

divide both sides by 2 we get;

or

the related equation for the inequality equation [1] is;

y-intercepts defined as the graph crosses the y-axis.

substitute x =0 in [2] to solve for y;

y = 6

x-intercepts defined as the graph crosses the x-axis.

substitute value y = 0 to solve for x;

subtract 6 from both sides we get

multiply both sides by 2 we get;

or

x = -12

now, plot these point (0, 6) and (-12, 0) on graph as shown below and shade the graph below the solid line.

parallelogram is the   answer