## Math 8 Chapter 1 Lesson 9: Rectangle

## 1. Theoretical Summary

### 1.1. Define

**Define:** A rectangle is a quadrilateral with four right angles.

**Eg:** Rectangle ABCD has 4 right angles.

From this definition, we deduce:

- A rectangle is an isosceles trapezoid with a right angle.
- A rectangle is a parallelogram with a right angle.

### 1.2. Nature

– Since a rectangle is an isosceles trapezoid and also a parallelogram, it has the properties of an isosceles trapezoid and the properties of a parallelogram, especially:

- In a rectangle, two diagonals are equal and intersect at the midpoint of each.
- Conversely, if a quadrilateral has two equal diagonals and intersects each other at the midpoint of each line, then the quadrilateral is a rectangle.

### 1.3. Center of symmetry – Axis of symmetry of the rectangle

– A rectangle has a center of symmetry that is the intersection of two diagonals.

– A rectangle has two axes of symmetry, which are two lines \({d_1},{d_2}\) passing through the midpoints of two opposite sides.

– The axes of symmetry of the rectangle pass through the center of symmetry, are perpendicular to the sides, and are perpendicular to each other.

### 1.4. Prove that a quadrilateral is a rectangle

To prove that a quadrilateral is a rectangle, we can show that it has one of the following four properties:

- There are three right angles
- Is an isosceles trapezoid with a right angle
- Is a parallelogram with a right angle
- There are two equal diagonals that intersect at the midpoint of each, or a parallelogram with two equal diagonals.

**– Theorem: **In a right triangle, the median on the hypotenuse is half of the hypotenuse. Conversely, in a triangle, if the median from a vertex is half of the opposite side, then the triangle is a right triangle.

– This theorem is often used to prove that straight segments are congruent and the opposite part is used to prove a right triangle.

– Points that are a constant distance from a given line a h lie on two lines parallel to a and a distance equal to h from a.

**– Theorem:**

- If equidistant parallel lines intersect a line, then if they intercept on that line successive segments are congruent.
- If parallel lines intersect a line and they intercept on that line consecutive lines that are equal, then they are equidistant parallel.

## 2. Illustrated exercise

**Question 1: **Given an acute triangle ABC, the orthocenter H and the intersection of the perpendicular bisectors is the point O. Let P, Q, N, respectively, be the midpoints of the line segments AB, AH, AC.

a. Prove that quadrilateral OPQN is a parallelogram.

b. What condition must triangle ABC have for quadrilateral OPQN to be a rectangle?

__Solution guide__

a. O is the intersection of the orthogonal lines, so:

\(OP \bot AB;\,\,\,\,ON \bot AC\)

In \(\Delta AHC,\) QN is the moving average, so QN//HC.

Which \(HC \bot AB\) should \(QN \bot AB.\)

So OP // QN (1)

Similar proof, we have

ON // PQ (2)

(1) and (2) deduce dcm

b. Let OPQN be a rectangle then

\(PQ \bot QN \Rightarrow HB \bot HC.\)

Obviously in this case point H must coincide with point A, i.e. triangle ABC is right-angled at vertex A.

**Verse 2:** Given triangle ABC, vertex A; AD bisector. Through D draw a line parallel to AB, this line intersects side AC at point E. Through E we draw a line parallel to BC, line parallel to BC, this line intersects AB at point F.

a. Prove AE = BF

b. Determine the shape of triangle ABC if point E is the midpoint of side AC.

__Solution guide__

a. Quadrilateral BDEF is a parallelogram giving us

BF = ED (1)

\(DE = AB \Rightarrow \widehat {{D_1}} = \widehat {{A_1}}\)

Assume for \(\widehat {{A_2}} = \widehat {{A_1}}\)

So \(\widehat {{D_1}} = \widehat {{A_2}} \Rightarrow \Delta AED\)weigh

Deduce AE = ED (2)

From (1) and (2) deduce dcm

b. When E is the midpoint of AC, then \(DE = \frac{1}{2}AC\)

\( \Rightarrow \Delta ADC\) square at D or AD is the altitude of \(\Delta ABC.\) Suppose AD is the bisector of angle A. So \(\Delta ABC\) is isosceles at A.

**Question 3: **Given rectangle ABCD. From vertex B draw BH perpendicular to diagonal AC (H belongs to AC). Let M, N, P, Q be the midpoints of the line segments AH, AB, NC and DC, respectively.

a. Prove \(MP = \frac{1}{2}NC.\)

b. Prove \(BM \bot MQ\)

__Solution guide__

a. In \(\Delta ABH\), MN is the moving average: MN // BH

\( \Rightarrow \Delta NMC\) is square of vertex M, MP is the median of hypotenuse NC, so

\(MP = \frac{1}{2}NC\)

b. Quadrilateral BNQC is a rectangle; P is the intersection of the two diagonals; NC = BQ

Derive \(MP = \frac{1}{2}BQ\)

Triangle BMQ has median MP equal to half of the corresponding side BQ. So it’s a right triangle at the vertex M, so \(BM \bot MQ.\)

## 3. Practice

### 3.1. Essay exercises

**Question 1:** Let ABC be a triangle. From vertex A, we draw lines AP, AQ in order perpendicular to the exterior bisectors of angle B; The lines AR, AS are in the order perpendicular to the internal and external bisectors of angle C. Prove:

a. The quadrilaterals APBQ, ARCS are rectangles.

b. Four points Q, R, P, S are collinear.

c. \(QS = \frac{1}{2}(AB + BC + CA).\)

d. What conditions must be satisfied in triangle ABC for APBQ to be a square? From this it follows that it is impossible for both quadrilaterals APBQ and ARCS to be squares.

**Verse 2: **Given an isosceles triangle ABC, vertex A. From a point D on the base of BC we draw a line perpendicular to BC, which intersects AB at E and AC at point F. Draw the rectangles BDEH and CDFK. Let I, J be the centers of the rectangles BDEH and CDFK, respectively, and M be the midpoints of the line segment AD.

a. Prove that the midpoint of line segment HK is a fixed point, independent of the position of point D on side BC.

b. Prove that the three points I, M, J are collinear and the three lines AD, HJ, KI are concurrent.

c. When point D moves on side BC, on which line segment does point M move?

**Question 3:** Given triangle ABC, draw altitudes BE and CD and call H the orthocenter of the triangle. M, N, K are respectively the midpoints of the line segments BC, AH and DE.

a. Prove that the three points M, N, K are collinear

b. Based on the above results, it follows that if we call P, Q, R, S the midpoints of AB, HC, AC, HB, respectively, the three lines MN, PQ, RS are concurrent.

### 3.2. Multiple choice exercises

**Question 1: ** Which of the following signs is not correct?

A. A quadrilateral with 3 right angles is a rectangle

B. An isosceles trapezoid with a right angle is a rectangle

C. A parallelogram with a right angle is a rectangle

D. A parallelogram with two perpendicular diagonals is a rectangle

**Verse 2: **Let us, triangle ABC is right angled at A, AM is the median. We know BC = 10 cm. Ask AM =?

A. AM = 10 cm

B. AM = 5 cm

C. AM = 12 cm

D. Insufficient data to calculate AM

**Question 3: **Let ABCD be a rectangle with AB = 5 cm and the area of the rectangle is 40 cm2. What is the length of the diagonal?

A. \(\sqrt {89} \,\,\)cm

B. \(\sqrt {98} \,\)cm

C. 10 cm

D. 8 cm

**Question 4: ** Given a right triangle with the median to the hypotenuse of length 6 cm, what is the length of the hypotenuse?

A. 6 cm

B. 3 cm

C. 12 cm

D. 10 cm

**Question 5:** Choose the wrong idea

A. Rectangle has all properties of parallelogram and isosceles trapezoid

B. Rectangle whose center of symmetry is the intersection of two diagonals

C. A rectangle has two axes of symmetry

D. A rectangle is a quadrilateral with four right angles and four equal sides

## 4. Conclusion

Through this lesson, you should know the following:

- Recognize rectangles.
- Remember the properties and signs of rectangles.
- Apply knowledge to solve some related problems.

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