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Veronika [31]
2 years ago
9

8. In quadrilateral ABCD, AD is congruent to BC, and AD is parallel to BC. Andre has written a proof to show that ABCD is a para

llelogram. Fill in the blanks to complete the proof.
​

Mathematics
1 answer:
luda_lava [24]2 years ago
8 0

The correct choices to fill the blanks are listed below:

  1. AD
  2. BC
  3. AC
  4. CBA
  5. angle BAC
  6. CD

<h3>How to prove that a quadrilateral is a parallelogram</h3>

In this question we should fill the blanks based all information related to Euclidean geometry, especially concepts related to angles, triangles, parallelism and quadrilaterals.

The complete paragraph is shown below:

<em>Since </em><u><em>AD</em></u><em> is parallel to </em><u><em>BC</em></u><em>, alternate interior angles. </em>

<u><em>AD</em></u><em> and </em><u><em>BC</em></u><em> are congruent. </em>

<em>AC is congruent to </em><u><em>AC</em></u><em> since segments are congruent to themselves.</em>

<em>Along with the given information that AD is congruent to BC, triangle ADC is congruent to triangle </em><u><em>CBA</em></u><em> by the Side-Angle-Side Triangle Congruence. </em>

<em>Since the triangles are congruent, all pairs of corresponding angles are congruent, so angle DCA is congruent to </em><u><em>angle BAC</em></u><em>.</em>

<em>Since those alternate interior angles are congruent. AB must be parallel to </em><u><em>CD</em></u><em>. </em>

<em>Since we define a parallelogram as a quadrilateral with both pairs of opposite sides parallel, ABCD is a parallelogram.</em>

To learn more on quadrilaterals, we kindly invite to check this verified question: brainly.com/question/25240753

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<u>Step-by-step explanation:</u>

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Of the given options, only A & D are candidates.  Now we need to decide if it is a cos graph (A) or a sin graph (D).  If we shift the graph up 5 units (to eliminate the vertical shift) , the graph would pass through the origin.  Thus it is a sin graph and the answer must be option D.

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Please help: linear algebra problem. (Linear combinations)
DochEvi [55]

Answer:

\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]

This tells us that:

A=\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right]

b=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]

Step-by-step explanation:

So we are saying we have scalars, c and d, such that:

c\left[\begin{array}{ccc}5\\5\\ 3\end{array}\right]+d\left[\begin{array}{ccc}7\\-8\\-9\end{array}\right]=\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right].

So we want to find a way to express this as:

Ax=b where x is the scalar vector, \left[\begin{array}{ccc}c\\d\end{array}\right].

So we can write this as:

\left[\begin{array}{ccc}5&7\\5&-8\\3&-9\end{array}\right] \left[\begin{array}{ccc}c\\d\end{array}\right] =\left[\begin{array}{ccc}-16\\3\\-15\end{array}\right]

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