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Setler [38]
3 years ago
15

Prove that opposite angles of a parallelogram are congruent. Be sure to create and name the appropriate geometric figures.

Mathematics
1 answer:
qwelly [4]3 years ago
7 0
<span>A parallelogram is a quadrilateral whose opposite sides are parallel. ABCD is a parallelogram; AB II CD and AC II BD

Theorem: </span><span>opposite angles of a parallelogram are congruent
Given: Parallelogram PQRS
Prove: </span>∠CAB = ∠BDC and ∠ABD= ∠DCA

Refer to the attached picture to follow.

Create an extended line beyond and name it as point E and point F.
Proof:
   ∠CDB = ∠EBD - alternate interior angle
   ∠EBD = ∠CAB - corresponding angle
      This makes ∠CDB congruent to ∠CAB.
   ∠ACD = ∠FAC - alternate interior angle
   ∠FAC = ∠ABD - corresponding angle
      This makes ∠ACD congruent to ∠ABD

Proving that ∠CAB = ∠BDC and ∠ABD= ∠DCA or ∠CAB = ∠CDB and ∠ABD = ∠ACD

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Answer:

14/20 or .7 or 70%

Step-by-step explanation:

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What must be true for lines a and b to be parallel lines? Check all that apply.
Dima020 [189]

Answer:

Option (1), (2) , (7) and (8) is correct.

Step-by-step explanation:

Given: a║b  and d and c be transversal , We have to find the value of x and measure of angle 1 and 2.

Since, a║b  and d be a transversal  then ,

∠BXZ = ∠AZP = 58° ( corresponding angles )

Thus, m∠2 = 58°

∠BYC = ∠XYZ = (4X-10)° (vertically opposite angles)

Since, a║b  and c be a transversal  then,

  ∠XYZ = ∠QZU = (4X-10)°  (Corresponding angles)

Thus, m∠1 =  (4X-10)°

Also, Since, 'a' is a straight line.

Sum of angles on a straight line is 180°.

⇒ ∠PZA + ∠PZQ + ∠QZU = 180°

⇒ 58° +(3X-1)°+(4X-10)° = 180°

⇒ 58° + 7x - 11 = 180°

⇒ 47° + 7x  = 180°

⇒  7x  = 180°- 47°

⇒  7x  = 133°

⇒  x  = 19°

Put x = 19 in (3X-1)° and (4X-10)° , we get

(3X-1)° = (3(19)-1)° = 56°

(4X-10)° = (4(19)-10)°= 66

Thus, (3X-1)° ≠ (4X-10)°

Thus, option (1), (2) , (7) and (8) is correct.


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