# Vertices Of A Parallelogram Assignment Definition

## Answers for 4C

### 4.3 Midpoint Quadrilaterals (6 points)

If ABCD is a quadrilateral, then * its midpoint quadrilateral MNOP* is the quadrilateral with vertices at the midpoints of the sides: M = midpoint AB, N = midpoint BC, etc.

- For a quadrilateral ABCD, prove that the midpoint quadrilateral of ABCD is a parallelogram.

Answers: Here are some student answers, *all of them good*, all different, but some open to discussion about possible ways to be more concise. In each case, ask

What criterion for parallelogram is used here?

Is it necessary to repeat so many steps? If you state what you prove as something like "any side of the midpoint quadrilateral is parallel to the opposite side", then would this make it possible to make the argument only once?

Answer A: | Answer B | Answer C

**4.4 Special Midpoint Quadrilaterals (10 points)**

Prove, if true. Construct a counterexample if false.

- If ABCD is a kite, then its midpoint quadrilateral is a rectangle.
- If the midpoint quadrilateral of ABCD is a rectangle, then ABCD is a kite.
- If ABCD is a rectangle, then its midpoint quadrilateral is a rhombus.
- If the midpoint quadrilateral of ABCD is a rhombus, then ABCD is a rectangle.

ANSWER

### 4.5 Median Intersection (10 points)

Draw a triangle ABC and construct midpoints D, E, F as in the figure. Let G be the intersection of the two medians CF and BD. Then let H be the midpoint of BG and I be the midpoint of CG.

- Prove that DFHI is a parallelogram.

DFHI is the midpoint quadrilateral of CABG, so this follows from 4.3.

- Then use what you know about parallelograms to find the ratio BG/BD and also CG/CF.

The diagonals of the parallelgram DFHI bisect each other, so DG = HG. Also, since H is midpoint of BG, HG = BH. Since BD = DH + HG + GD, BG/BD = 2GD/3GD = 2/3.

The same argument applies to any median intersecting another median, so CG/CF = 2/3 also.

- Finally, repeat this construction for medians BD and AE, intersecting in J. Can you prove G = J?

Repeating the same reasoning BJ/BD = AJ/AE = 2./3.

But this says that BJ = (2/3)BD = BG. But there is only one point in segment BD whose distance from B is (2/3)BD, so the two points G and J must be the same, and all the medians are concurrent at G.

**4.6 Two-edged ruler figure (6 points)**

Suppose you place your ruler on paper and without moving it, draw lines along both edges of the ruler. Then turn the ruler (no special angle) and draw lines along both edges again.

- Prove that the quadrilateral formed by the 4 lines is a rhombus.

More formally, there are given two parallel lines m1 and m2 which are distance d apart. Let n1 and n2 be another pair of parallel lines that are the same distance d apart, but which are not parallel to the first pair. Then the lines will intersect at four points, forming a parallelogram (by definition!). Prove that this parallelogram is a rhombus.

Student Answer (correct). Note that the standard mistake on this problem is to confuse the distance between lines (given as equal) with the distances between vertices (to be proved equal).

**4.7 Circumcenter of an Isosceles Triangle (8 points)**

Given an isosceles triangle ABC, with |AB| = |AC| = b and |BC| = a, what is the radius R of the circumcircle of ABC? (The answer should be in terms of a and b, if possible.)

You can start off this problem by doing a sketch of the parallelogram, just to have a visual representation.

Secondly, since you need to find the perimeter and area of the parallelogram, that requires knowing the lengths of the sides. Use the distance formula for all sides.

√ = square root, x_{1} = 1st x- coordinate, x_{2} = 2nd x-coordinate, y_{1}=1st y- coordinate, and y_{2} = 2nd y-coordinate

all values are under the square root sign

√(x_{2}-x_{1})² + (y_{2}-y_{1})² = distance formula

so I assigned letters to each coordinate: A(0,3), B(3, 0), C(0, -3) and D(-3, 0)

AB: √(3 - 0)^{2} + (0 - 3)^{2} = √(3)^{2} + (-3)^{2} = √(9) + (9) = √18

BC: √(0 - 3)² +(-3 - 0)² = √(-3)² +(-3)² =√(9) + ( 9) = √18

CD: √(-3 - 0)² +(0 - -3)² = √(-3)² +(3)² = √(9) + (9) = √18

AD: √(-3 - 0)² +(0 - 3)² = √(-3)² +(-3)² =√(9) + (9) = √18

Since all sides are congruent,therefore Its a square.

Area = s^{2} = √18² = 18

Perimeter = add all sides: 4(√18)= 4√18, but in simplest radical form its: 4• √9 •√2 = 4 • 3 •√2 = 12√2