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25 Dec 2020, 21:13
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I wish I could draw a good picture, but hopefully something will be useful:
To show that whenever you have the 2 Diagonals of a Quadrilateral Bisect Each other, the Quadrilateral must be a Parallelogram:
-you can draw a 4 Sided Quadrilateral with the 2 Diagonals Bisecting Each Other
-the 2 Diagonals Bisecting Each Other creates 4 Triangles within the Quadrilateral
-using the Side-Angle-Side Theorem to prove Triangle Congruency, you can show how the 2 Pairs of Vertically Opposite Triangle are Congruent. If 2 Sides and the Included Angle of a Triangle are Congruent to the 2 Corresponding Sides and Included Angle of another Triangle, then those 2 Triangles are Congruent. Use the 1/2 of Each of the Bisected Diagonals and the Included Angle in between for each Triangle to show how the corresponding parts are congruent.
-because there is 2 Pairs of Congruent Triangles within the Quadrilateral, for Each Pair of Congruent Triangles the Opposite Sides of the Quadrilateral are equal, corresponding Sides. This shows that the Opposite Sides of the Quadrilateral are Equal
-and finally, if a Quadrilateral has 2 Pairs of Congruent, Opposite Sides ------> it must be a Parallelogram
-then, add in Statement 2, in which 1 Angle is 90 degrees
-if 1 angle of a Parallelogram is 90 Degrees, then the other 3 Angles will also be 90 Degrees (Can use Either of 2 Rules: Adjacent Angles of a Parallelogram are Supplementary or Vertically Opposite Angles of a Parallelogram are Equal)
So, we have Opposite Sides must Equal in this Parallelogram and each Interior Angle is 90 degrees. It must be a Rectangle
(C)
Probably easier to remember:
Theorem: If a Quadrilateral is a Parallelogram, then the 2 Diagonals will Bisect Each Other
Converse of Theorem is also True: If a Quadrilateral has its 2 Diagonals Bisect Each Other, then the Quadrilateral must be a Parallelogram.
and
In any Parallelogram, Adjacent Angles are Supplementary (add up to 180 degrees)
and
In any Parallelogram, Vertically Opposite Angles are Congruent.
To show that whenever you have the 2 Diagonals of a Quadrilateral Bisect Each other, the Quadrilateral must be a Parallelogram:
-you can draw a 4 Sided Quadrilateral with the 2 Diagonals Bisecting Each Other
-the 2 Diagonals Bisecting Each Other creates 4 Triangles within the Quadrilateral
-using the Side-Angle-Side Theorem to prove Triangle Congruency, you can show how the 2 Pairs of Vertically Opposite Triangle are Congruent. If 2 Sides and the Included Angle of a Triangle are Congruent to the 2 Corresponding Sides and Included Angle of another Triangle, then those 2 Triangles are Congruent. Use the 1/2 of Each of the Bisected Diagonals and the Included Angle in between for each Triangle to show how the corresponding parts are congruent.
-because there is 2 Pairs of Congruent Triangles within the Quadrilateral, for Each Pair of Congruent Triangles the Opposite Sides of the Quadrilateral are equal, corresponding Sides. This shows that the Opposite Sides of the Quadrilateral are Equal
-and finally, if a Quadrilateral has 2 Pairs of Congruent, Opposite Sides ------> it must be a Parallelogram
-then, add in Statement 2, in which 1 Angle is 90 degrees
-if 1 angle of a Parallelogram is 90 Degrees, then the other 3 Angles will also be 90 Degrees (Can use Either of 2 Rules: Adjacent Angles of a Parallelogram are Supplementary or Vertically Opposite Angles of a Parallelogram are Equal)
So, we have Opposite Sides must Equal in this Parallelogram and each Interior Angle is 90 degrees. It must be a Rectangle
(C)
Probably easier to remember:
Theorem: If a Quadrilateral is a Parallelogram, then the 2 Diagonals will Bisect Each Other
Converse of Theorem is also True: If a Quadrilateral has its 2 Diagonals Bisect Each Other, then the Quadrilateral must be a Parallelogram.
and
In any Parallelogram, Adjacent Angles are Supplementary (add up to 180 degrees)
and
In any Parallelogram, Vertically Opposite Angles are Congruent.
19 Feb 2021, 06:54
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VeritasKarishma plz solve this
21 Feb 2021, 23:27
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gottabwise
Recall how we go from general quadrilateral to specific:
Quadrilateral
- trapezoid (one set of opposite sides parallel)
- parallelogram (both sets of opposite sides parallel, diagonals bisect each other)
- rhombus (all sides equal and opposite sides parallel, diagonal bisect and perpendicular)
- rectangle (Opposite sides parallel and all 90 degree angles, diagonals bisect)
- square (all sides equal, all 90 degree angles, diagonals bisect and perpendicular)
(1) Line segments AC and BD bisect one another.
All we can say is that we have a parallelogram at least. Is it a rectangle, we don't know.
(2) Angle ABC is a right angle.
We have one right angle. It's possible that all other 3 angles are not 90 degrees. Not sufficient.
Using both, we have a parallelogram with a 90 degree angle. So the angle opposite to 90 degree angle will also be 90 degrees. This means the other two angles with have a sum of 180 and will be equal so then they both will be 90 too. Hence all angles will be 90.
A parallelogram with all angles 90 degrees is a rectangle.
Sufficient
Answer (C)
