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: "Jeannine Ramacho" <jeannine_ramacho@morris.lakeside.sea.wa.us> Winners
James and Kairu, Lakeside High School, Seattle, Washington
The question of the month was, what shape is formed when the midpoints of a quadrilateral are connected. To answer this question we have decided to start with a quadrilateral with no specific properties. To prove that quadrilateral PQRS is a parallelogram we added AC to quadrilateral ABCD. Because the midpoints of triangle ABC are connected the resulting triangle PBQ is similar to ABC and exactly half the size, in addition to that the lines PQ and AC are parallel. The triangle ACD has the midpoints R and S connected resulting in the similar triangle SDR which is exactly half the size of triangle ACD, and in addition to that the lines SR and AC are parallel. Therefore since SR is parallel to and exactly one half of AC, and PQ is also parallel to and exactly one half AC. Therefore by the transitive property PQ is equal and parallel to RS. Therefore quadrilateral PQRS is a parallelogram because it has two equal parallel segments. Therefore all quadrilaterals that have their consecutive midpoints connected form a parallelogram.
Quadrilateral ABCD is a rectangle, therefore AD is equal to BC, and since S and Q bisect AD and BC, respectively, SD is equal to QC, is equal to BQ is equal to AS. AB is equal to DC because ABCD is a rectangle, and P and R bisect AB and DC. Therefore DR is equal to RC is equal to BP is equal to PA because they are all one half of equal segments. Angles A, B, C, and D are all equal to 90 degrees because quadrilateral ABCD is a rectangle. Therefore triangles APS, BPQ, CRQ, and DRS are all congruent by side angle side. Since those triangles are equal the sides QR, RS, SP, and PQ are all equal because they are corresponding parts of congruent triangles. Therefore quadrilateral PQRS is a rhombus because it is constructed of four equal sides.
This is a square. Therefore, it has all the qualities of a parallelogram plus two more. All sides are equal and all angles equal 90º. If you connect the consecutive midpoints of each side, they form a square and four right isosceles triangles. The triangles have one right angle, and since all segment are equal, the bisected segments will be equal as well, so the right triangle will have two equal sides. B'AIT are equal. Therefore, all triangles have angles equal 90 degrees - 45 degrees - 45 degrees.
PQRS is a square because it is formed by the four hypotenuses of four right isosceles triangles. The adjacent angles to a hypotenuse would in this case equal 45 degrees. Therefore the supplement, the angle of the inner square, equals 90š, and therefore the inside quadrilateral is a square.
Quadrilateral ABCD is a rhombus, therefore angle A equals angle C, angle B equals angle D, and all sides are equal. Points P, Q, R, and S are midpoints which forces AS=SD=DR=RC=CQ=QB=BP=PA because they are all one half of equal segments. Triangles QRC and APS are congruent by side angle side. Triangles BPQ and DRS are also congruent by side angle side. Angle SRQ is equal to angle to RQP, is equal to QPS, is equal to PSR. The reason for this is angles RSD, SRD, BQP, and BPQ are all equal because they are base angles of identical isosceles triangles. Angles CRQ, CQR, APS, and ASP are also equal because they are base angles of identical isosceles triangles. Therefore because CQR+BQP is supplementary to angle PQR, CRQ+SRD is supplementary to SRQ, RSD+ASP is supplementary to angle PSR, and APS+BPQ is supplementary to SPQ, angles PQR, QRS, RSP, and SPR are equal because all their supplements (CQR+BQP, CRQ+SRD, RSD+ASP, APS+BPQ) are all equal. Therefore since there are four equal angles in a quadrilateral and there are only 360 degrees in any quadrilateral each angle must equal 90 degrees. Therefore quadrilateral PQRS is a rectangle because it has four angles equal to 90 degrees. Therefore this is a rectangle, because there are two pairs of equal parallel sides, and all angles are right angles.
Quadrilateral ABCD is an isosceles trapezoid, and PQRS is a parallelogram because the points PQRS are midpoints, and I proved at the beginning that any quadrilateral with the midpoints connected forms a parallelogram. AS is equal to BQ because they are each one half of equal segments. AP is equal to PB because P is the midpoint of AB. Angle A is equal to angle B because ABCD is an isosceles trapezoid, therefore triangles SAP, and PBQ are congruent by side angle side. Therefore SP is equal to PQ by corresponding parts of congruent triangles. Therefore PQRS is a rhombus because SP = PQ, and PQ = SR (because it is a parallelogram) and PS = QR (because it's a parallelogram.
Comments:
James and Kairu presented the general case and proved it, and then went on to prove five examples fairly thoroughly and correctly. The pictures included were very handy. Basically, they just did more parts correctly than anyone else. The main weakness in their solution is the general form - I am a sucker for things like introductions and conclusions, which they neglected to include. That is something for them to work on for next time.
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