What figure is formed? - December 1995
What figure is formed when the consecutive midpoints of the sides of a quadrilateral are joined? What if the original quadrilateral were a rectangle? A kite? An isosceles trapezoid? A square? A rhombus? Other shapes? Explain why you think your answer is true.
From: ruth@forum.swarthmore.edu (Ruth Carver) Honorable Mention #2
Katie Walder and Susan Tull
Grade 9
Mount St. Joseph AcademyThe quadrilateral's connected midpoints will form a parallelogram. We know that a quadrilateral is any figure with four sides. Drawing the diagonal of the quadrilateral, you form 2 triangles. After using the theorem that states a line which connects the midpoints of 2 sides of a triangle is parallel to the third side, you can then use the theorem that 2 lines parallel to a third line are parallel to each other. Do this for all the midpoints. You have then proved the figure is a parallelogram with 2 pairs of opposite parallel sides.
The rectangle would form a rhombus. A rectangle must have 4 ninety degree angles and two pairs of parallel congruent sides. You can prove this by proving all of the surrounding triangles are congruent. We do this by using the definition of a midpoint, the 90 degree angles, and the SAS postulate. This means that all of the lines of the parallelogram will be congruent by CPCTC.
A kite would form a parallelogram. (We are using the definition of a kite that states that it is a quadrilateral with 2 pairs of congruent sides, but opposite sides are not congruent. In other words, each pair of congruent sides are consecutive to each other.) First, we draw a line in the kite from the vertex between 2 non-congruent sides to the opposite vertex, forming 2 triangles. We know that the connecting lines of the midpoints would have to be two pairs of parallel lines, forming an isosceles trapezoid in each triangle. Since the two smaller bases must be 1/2 of the larger base, and the trapezoid shares the larger base, the 2 smaller bases are congruent. You can then draw the line connecting the 2 other vertices and perform the same operation. This would lead you to the same conclusion about the other set of lines. Or you could simply deduce this because a kite is a quadrilateral and we have already proven that a quad. forms a parallelogram.
An isosceles trapezoid will form a rhombus. By drawing a line through the midpoints of the bases it forms 2 triangles. By using the definition of a midpoint and the property that states the 2 top angles of an isosceles trapezoid are congruent you prove the 2 top surrounding triangles congruent. Since you already know the figure is a parallelogram, opposite sides are congruent. Therefore all 4 sides are congruent complying with the definition of a rhombus.
A square will form a square. Since all of a square's sides and angles are congruent, the lines that connect their midpoints must also be congruent. We proved this by using the SAS postulate to prove all of the surrounding triangles are congruent.
A rhombus will form a parallelogram. We know this because we have already proven that a quadrilateral will form a parallelogram.
All of the following shapes plus others will form themselves when their midpoints are connected: triangle, pentagon, hexagon, a seven-sided figure, an octagon, a nonagon, and so on, as well as any other n-gon that does not have 1, 2, or 4 sides. The quadrilateral works as long as you leave the figure general and do not look beyond it having four sides. A square forms a square. However, a rectangle does not form a rectangle, it forms a rhombus.
Comments:
A very thorough job. Pretty good proofs of the examples. The presentation could be a little bit better, but the real problem here is that they got a couple of the examples wrong! In both cases where you get a rectangle, the rhombus and the kite, they stopped at parallelogram. A little bit more investigation might have led them to notice the 90 degree angles. The rest are really nice.
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