## Math 8 Chapter 1 Lesson 7: Parallelogram

## 1. Theoretical Summary

### 1.1 Definitions

A parallelogram is a quadrilateral whose opposite sides are parallel.

ABCD is a parallelogram \( \Leftrightarrow \) AB // CD and AD // BC.

Thus, a parallelogram is a trapezoid with two parallel sides.

### 1.2 Properties

Theorem: In a parallelogram then:

a) Opposite sides are equal.

ABCD is a parallelogram \( \Rightarrow \) AB =DC and AD = BC.

b) Opposite angles are congruent

ABCD is a parallelogram \( \Rightarrow \)\(\widehat {A\,} = \widehat C\) and \(\widehat B = \widehat D\)

c) Two diagonals intersect at the midpoint of each line.

ABCD is a parallelogram \( \Rightarrow \) OA = OC and OB = OD

### 1.3. Recognizing signs

A quadrilateral with one of the following properties is a parallelogram

1. Opposite sides are parallel

Quadrilateral ABCD has AB // CD and AD // BC \( \Rightarrow \) ABCD is a parallelogram.

2. Opposite sides

Quadrilateral ABCD has AB = CD and AD = BC \( \Rightarrow \) ABCD is a parallelogram.

3. Opposite angles are equal

Quadrilateral ABCD has \(\widehat {A\,} = \widehat C\) and \(\widehat B = \widehat D\) \( \Rightarrow \) ABCD is a parallelogram.

4. There are two diagonal lines that intersect at the midpoint of each line.

Quadrilateral ABCD has OA = OC and OB = OD \( \Rightarrow \) ABCD is a parallelogram.

5. There are two opposite sides parallel and equal

Quadrilateral ABCD has AB // CD and AB = CD or AD // BC and AD = BC \( \Rightarrow \) ABCD is a parallelogram.

## 2. Illustrated exercise

**Question 1: **Let ABCD be a parallelogram. On the side AB, BC, CD, DA in order, take the points M, N, P, Q such that AM = BN = CP = DQ. Prove that quadrilateral MNPQ is a parallelogram.

**Solution guide: **

ABCD is a parallelogram giving us:

AB = CD and \(\widehat B = \widehat D\)

According to the assumption AM = CP, from above we infer: BM = DP

Consider two triangles BMN and DPQ with:

\(\begin{array}{l}BM = DP\\\widehat B = \widehat D\\BN = DQ\\ \Rightarrow \Delta BMN = \Delta DPQ \Rightarrow \widehat M = \widehat P\,\ ,\,\,\,\,\,\,\,(1)\end{array}\)

Let E be the intersection of PQ and the line AB .

\(DC//AB \Rightarrow \widehat P = \widehat E\) (2)

From (1), (2) and since \(\widehat M,\widehat E\) are two angles in isotopic positions, it follows that: MN // PQ

Similar proof we have: MQ // NP

The quadrilateral MNPQ has opposite sides parallel, so by definition it is a parallelogram.

**Verse 2**: Given an isosceles triangle ABC, AB = AC and P is any point on the base side BC. Let M, N, respectively, be the midpoints of the lines BP, CP. The perpendicular bisector of BP intersects side AB at point E. The perpendicular bisector of CP intersects side AC at point F.

a. Prove that quadrilateral AEPF is a parallelogram

b. The sum of PE + PF does not depend on the choice of point P on BC.

**Solution guide: **

a. \(\Delta PFC\) equals vertex F because FP = FC

So \(\widehat {{P_1}} = \widehat C\) but \(\widehat C = \widehat B\)

\( \Rightarrow \widehat {{P_1}} = \widehat B \Rightarrow PF//AB\)

Similarly, we have PE // AC

Quadrilateral AEPF has opposite sides parallel.

So it is a parallelogram.

b. AEPF is a parallelogram

so PE = BE

PF = EA

\(\Rightarrow PE + PF = BE + EA = AB\)

The sum of PE + PF is always equal to the side AB of the isosceles triangle.

So it does not depend on the choice of point P on side BC.

**Question 3:** Let ABCD be a parallelogram. On the diagonal AC take two points E, F such that AE = EF = FC.

a. Prove that quadrilateral BEDF is a parallelogram

b. DF intersects BC at M. Prove DF = 2FM

c. BF intersects DC at I and DE intersects AB at J. Prove that the three points I, O, J are collinear.

**Solution guide: **

a. Let O be the intersection of the two diagonals, we have:

OA = OC

Combined with AE = CF, we get

OE = OF

\( \Rightarrow \) O is the midpoint of EF

We also have O as the midpoint of DB.

So BEDF is a parallelogram

b. We have \({\rm{OF}} = \frac{1}{2}{\rm{EF}} \Rightarrow {\rm{FC = }}\frac{2}{3}OC.\) triangle CDB, point F lies on median CO and a distance from the vertex by \(\frac{2}{3}\) median. So F is the centroid of \(\Delta CDB,\) so: DF = 2FM

c. F is the centroid of \(\Delta CDB\) so I is the midpoint of DC:

\(DI = \frac{1}{2}DC.\)

Similarly, E is the centroid of \(\Delta ADB\), so

\(BJ = \frac{1}{2}AB\)

So DI = BJ

Quadrilateral BIDJ has DI // BJ and DI = BJ so it is a parallelogram, so IJ passes through the midpoint O of DB.

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Let ABC be a triangle. Let D, E, F be the midpoints of the sides AB, AC, BC, and I, J, K respectively the midpoints of the lines DF, BF and CD. Prove:

a. The quadrilaterals IJFK and IEKJ are parallelograms

b. Three points E, K, F are collinear

**Verse 2: **Prove that:

a. In a parallelogram, the intersection of the diagonals coincides with the intersection of the line segments joining the midpoints of the opposite sides

b. Conversely, if a quadrilateral has the intersection of two diagonals coincident with the intersection of the lines joining the midpoints of the opposite sides, then the quadrilateral is a parallelogram.

**Question 3:** Given quadrilateral ABCD. Let E, F, respectively, be the midpoints of two opposite sides AB, CD, M, N, P, Q, respectively, the midpoints of the lines AF, CE, BF and DE, respectively.

a. Prove that the lines MP, NQ and EF intersect at the midpoint of each line

b. The quadrilateral MNPQ is a parallelogram.

### 3.2. Multiple choice exercises

**Question 1: **Choose the wrong idea

A. A quadrilateral with opposite sides parallel is a parallelogram

B. A quadrilateral with equal opposite sides is a parallelogram

C. A quadrilateral with two equal opposite angles is a parallelogram

D. A quadrilateral with two diagonals intersecting at the midpoint of each is a parallelogram

**Verse 2: **Choose the correct idea

A. A parallelogram is a quadrilateral with two opposite sides parallel

B. A parallelogram is a quadrilateral with equal angles

C. A parallelogram is a quadrilateral with opposite sides parallel

D. A parallelogram is a trapezoid with two equal sides

**Question 3: **Chop parallelogram ABCD with \(\angle A = {120^0}\) what are the remaining angles?

A. \(\angle B = {60^0}\angle C = {120^0}\angle D = {60^0}\)

B. \(\angle B = {110^0}\angle C = {80^0}\angle D = {60^0}\)

C. \(\angle B = {80^0}\angle C = {120^0}\angle D = {80^0}\)

D. \(\angle B = {120^0}\angle C = {60^0}\angle D = {120^0}\)

**Question 4: **Let parallelogram ABCD know \(\angle A – \angle B = {20^0}\) determine the measure of angle A and B

A. \(\angle A = {80^0}\angle B = {100^0}\)

B. \(\angle A = {100^0}\angle B = {80^0}\)

C. \(\angle A = {80^0}\angle B = {60^0}\)

D. \(\angle A = {120^0}\angle B = {100^0}\)

**Question 5: **Given parallelogram ABCD, let I be the intersection of the diagonals AC and BD. Choose the correct idea

A. AC = BD

B. Triangle ABD is isosceles at A

C. BI is the median of triangle ABC

D. \(\angle A + \angle C = \angle B + \angle D\)

## 4. Conclusion

Through this parallelogram lesson, students need to complete some of the objectives given by the lesson, such as:

- Understand the concept, properties, and signs of parallelograms
- Prove that a quadrilateral is a parallelogram
- Apply knowledge to solve some related problems

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