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Feliz [49]
4 years ago
7

What is the missing reason in the proof?

Mathematics
2 answers:
zhuklara [117]4 years ago
8 0

Answer:

Option 2.

Step-by-step explanation:

Given information: ABCD is a parallelogram with diagonal BD.

Prove: ΔABD ≅ΔCDB

Proof:

Definition of parallelogram: Opposite sides of a parallelogram are parallel and congruent.

Alternate interior angle theorem : If a transversal line intersect two parallel lines, then the alternate interior angles are congruent.

ASA postulate: If two corresponding angles and their included side are congruent, then both triangles are congruent.

Statement                                     Reason

1. AD║BD                                 1. Definition of parallelogram.

2. ADB≅CBD                           2. Alternate interior angle theorem.

3. AB║CD                                 3. Definition of parallelogram.

4. ABD≅CDB                           4. Alternate interior angle theorem.

5. DB≅DB                                5. Reflexive property of congruence

6. ΔABD ≅ΔCDB                    6. ASA

The missing reason is "Definition of parallelogram".

Therefore, the correct option is 2.

djverab [1.8K]4 years ago
3 0

the answeris segment addition


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Answer:

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Step-by-step explanation:

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<em>Additional comment</em>

Solving a compound inequality is very much like solving a single inequality. You need to "undo" what is done to the variable. The rules of equality (ordering) still apply. If you were to multiply or divide by a negative number, the direction (sense) of the inequality symbols would reverse in the same way they do for a single inequality.

Here, our first step was to subtract 2 from all parts of the inequality:

  38 -2 < 12x +2 -2 < 74 -2   ⇒   36 < 12x < 72

The division by 12 worked the same way: all parts are divided by 12.

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If it makes you more comfortable, you can treat the perimeter limits as two separate inequalities: 38 < 12x+2 and 12x+2 < 74. Both restrictions apply, so the solution set is the intersection of the solution sets of these separate inequalities.

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