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ioda
3 years ago
15

Since ABCD is a parallelogram, the two pairs of _________ sides (AB¯ and CD¯, as well as AD¯ and BC¯) are congruent. Then, since

∠9 and ∠11 are vertical angles, it can be concluded that ∠9≅∠11. Since ABCD is a parallelogram, AB¯∥CD¯. Since ∠2 and ∠5 are alternate interior angles along these parallel lines, the Alternate Interior Angles Theorem allows that ∠2≅∠5. Since two angles of △AEB are congruent to two angles of △CED, the Third Angles Theorem supports that ∠8≅∠3. Therefore, using the _________, it can be stated that △AEB≅△CED. Then, applying the definition of congruent triangles, it can be stated that AE¯≅CE¯, which makes E the midpoint of AC¯. Use a similar argument to prove that △AED≅△CEB; then it can be concluded that E is also the midpoint of BD¯. Since the midpoint of both line segments is the same point, the segments bisect each other by definition. Match each number (1 and 2) with the word or phrase that correctly fills in the corresponding blank in the proof.
Mathematics
1 answer:
Marysya12 [62]3 years ago
7 0

Answer:

1. Opposite

2. angle-side-angle criterion

Step-by-step explanation:

Since ABCD is a parallelogram, the two pairs of <u>(opposite)</u> sides (AB¯ and CD¯, as well as AD¯ and BC¯) are congruent. Then, since ∠9 and ∠11 are vertical angles, it can be concluded that ∠9≅∠11. Since ABCD is a parallelogram, AB¯∥CD¯. Since ∠2 and ∠5 are alternate interior angles along these parallel lines, the Alternate Interior Angles Theorem allows that ∠2≅∠5. Since two angles of △AEB are congruent to two angles of △CED, the Third Angles Theorem supports that ∠8≅∠3. Therefore, using the <u>(angle-side-angle criterion)</u>, it can be stated that △AEB≅△CED. Then, applying the definition of congruent triangles, it can be stated that AE¯≅CE¯, which makes E the midpoint of AC¯. Use a similar argument to prove that △AED≅△CEB; then it can be concluded that E is also the midpoint of BD¯. Since the midpoint of both line segments is the same point, the segments bisect each other by definition. Match each number (1 and 2) with the word or phrase that correctly fills in the corresponding blank in the proof.

A parallelogram posses the following features:

   1. The opposite sides are parallel.

   2. The opposite sides are congruent.

   3. It has supplementary consecutive angles.

   4. The diagonals bisect each other.

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1. Prove or give a counterexample for the following statements: a) If ff: AA → BB is an injective function and bb ∈ BB, then |ff
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a) False. A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1

b) True

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d) B = {1}, A = N, f: N ⇒ {1}, f(x) = 1

Step-by-step explanation:

a) lets use A = {1}, B = {1,2} f: A ⇒ B, f(1) = 1. Here f is injective but 2 is an element of b and |f−¹({b})| = 0., not 1. This statement is False.

b) This is True. If  A were finite, then it can only be bijective with another finite set with equal cardinal, therefore, B should be finite (and with equal cardinal). If A were not finite but countable, then there should exist a bijection g: N ⇒ A, where N is the set of natural numbers. Note that f o g : N ⇒ B is a bijection because it is composition of bijections. This, B should be countable. This statement is True.

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b) This is false, B = {1} is finite, A = N is infinite, however if f: N ⇒ {1}, f(x) = 1 for any natural number x, then f is surjective despite A not being finite.

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