Math 8 Chapter 1 Lesson 12: Square
1. Theoretical Summary
1.1. Define
A square is a quadrilateral with four right angles and four equal sides.
From this definition we deduce:
A square is a rectangle with two congruent adjacent sides.
A square is a rhombus with a right angle.
1.2. Nature
Since a square is both a rectangle and a rhombus, it has all the properties of a rectangle (For example, two diagonals are equal and intersect at the midpoint of each) and have all the properties of a rectangle. properties of a rhombus (e.g. two diagonals are perpendicular and each diagonal is the bisector of the vertices), in particular we can state the theorem:
In a square, two equal diagonals are perpendicular to each other and intersect at the midpoint of each.
1.3. Center of symmetry and axis of symmetry of the square
– A square has a center of symmetry that is the intersection of two diagonals.
– A square has four axes of symmetry, two diagonals and two lines connecting the midpoints of opposite sides.
1.4. Recognizing signs
To prove a square, we prove:
A rectangle has two congruent adjacent sides.
A rectangle has two diagonals that are perpendicular to each other.
– A rectangle whose diagonal is the bisector of an angle
– A rhombus has a right angle.
– A rhombus has two equal diagonals.
2. Illustrated exercise
Question 1: On a line xy, take three points A, B, C in the order of bro and AB > BC. In the same halfplane of the shore is the line xy, draw the squares ABDE and BCFG. A point H lies on segment AB and a point K lies on the opposite ray of ray DB such that AH = DK = BG. Prove that EHFK is a square.
Solution guide:
We have: \(\Delta EAK = \Delta KGF = \Delta HCF = \Delta EAH\)
\( \Rightarrow EK = KF = FH = HE\)
\( \Rightarrow \) EHFK is a rhombus (1)
\(\Delta EDK = \Delta KGF \Rightarrow \widehat {{E_1}} = \widehat {{K_1}}\)
Which \(\widehat {{E_1}} = \widehat {{K_2}} = {90^0} \Rightarrow \widehat {{K_1}} + \widehat {{K_2}} = {90^0}\)
\( \Rightarrow \widehat {EKF} = {90^0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\)
From (1) and (2) deduce dcm.
Verse 2: Given triangle ABC, perpendicular at vertex A, draw altitude AH and median AM. The bisector of angle A intersects the perpendicular bisector of side BC at point D. From D draw DE perpendicular to BC and DF perpendicular to AC. Prove:
a. AD is the bisector of angle HAM.
b. Three points E, M, F are collinear.
c. Triangle BDC is an isosceles right triangle.
Solution guide:
a. We have:
\(\widehat {{C_1}} = \widehat {{A_1}}\) (with \(\widehat B\))
\(\widehat {{C_1}} = \widehat {{A_2}}\) (\(\Delta AMC\)weight)
Derive \(\widehat {{A_3}} + \widehat {{A_4}}\)
\( \Rightarrow \) AD is the bisector of angle HAM.
b. AH // DM \( \Rightarrow \widehat {{D_1}} = \widehat {{A_4}}\) where \(\widehat {{A_4}} = \widehat {{A_3}}\)
\( \Rightarrow \widehat {{D_1}} = \widehat {{A_3}} \Rightarrow \Delta AMD\) balance
\( \Rightarrow \) MA = MD.
We can easily prove that AEDF is a square, giving
EA = ED; FA = FD
Thus, all three points E, M, F lie on the perpendicular bisector of the line segment AD.
c. \(\Delta BED = \Delta CFD \Rightarrow \widehat {{D_2}} = \widehat {{D_3}}\)
\(\widehat {BDC} = \widehat {BDF} + \widehat {{D_3}} = \widehat {BDF} + \widehat {{D_2}} = \widehat {EDF} = {90^0}\)
Combined with DB = DC deduce dcm.
Question 3: Let ABCD square. From a point M on the diagonal BD, draw ME perpendicular to AB and MF perpendicular to AD. Prove:
a. DE = CF and \(DE \bot CF.\)
b. CM = EF and \(CM \bot {\rm{EF}}{\rm{.}}\)
c. BF = CE and \(BF \bot CE.\)
d. From the above results, what can be concluded about the relative positions of the lines DE, CM and BF?
Solution guide:
a. \(\Delta EAD = \Delta FDC\)
\( \Rightarrow DE = CF\) and \(\widehat {{C_1}} = \widehat {{D_1}}\)
We have \(\widehat {{C_1}} + \widehat {{F_1}} = {90^0}\)
\(\begin{array}{l} \Rightarrow \widehat {{D_1}} + \widehat {{F_1}} = {90^0}\\ \Rightarrow \widehat {DKF} = {90^0}\\ \Rightarrow DE \bot CF.\end{array}\)
b. ABCD is a square; \(M \in BD\)
\( \Rightarrow \) MA = MC.
AEMF is the rectangle \( \Rightarrow \) MA = EF. So CM = EF.
\(\Delta CNM = \Delta EAF \Rightarrow \widehat {{C_2}} = \widehat {AEF}\)
\(AB//FN \Rightarrow \widehat {EKM} = \widehat {{M_1}}\) (isotope)
\(\widehat {EKM} + \widehat {AEF} = \widehat {{C_2}} + \widehat {{M_1}} = {90^0} \Rightarrow \widehat {KIE} = {90^0} \Rightarrow CM \bot {\rm{EF}}\)
c.
\(\begin{array}{l}\Delta B{\rm{AF = }}\Delta {\rm{CBE}} \Rightarrow BF = CE \Rightarrow \widehat {ABF} = \widehat {BCE}\\ \Rightarrow \widehat {BEC} + \widehat {ABF} = \widehat {BEC} + \widehat {BCE} = {90^0} \Rightarrow \widehat {EHB} = {90^0}\\ \Rightarrow BF \ bot CE.\end{array}\)
d. In triangle CEF, DE, CM and DF are altitudes, so they are concurrent.
Note: In this lesson, to prove that two lines are perpendicular, we show that they form with every third line a triangle whose sum of two angles is equal to \({90^0}\)
To prove that the three lines are concurrent, we show that they are special lines in the triangle (altitude, median, bisector, orthogonal).
3. Practice
3.1. Essay exercises
Question 1: Let ABC be a right triangle at vertex A and AB < AC. Line high AH. In the halfplane containing the vertex A, the border is the line BC, draw the figure AHDE.
a. Prove that point D belongs to the line segment HC.
b. Let F be the intersection of DE and AC. The line through F parallel to AB intersects the line through B parallel to AC at point G. Prove that quadrilateral ABGF is a square.
c. Prove that the three lines AG, BF and HE are concurrent.
d. Prove that quadrilateral DEHG is a trapezoid.
Verse 2: Let AB = a and any point M on that line. In the same half plane is the line AB, draw squares AMCD, BMEF.
a. Prove at a point we call H.
b. Prove that D, H, F are collinear.
c. Let I be the midpoint of line segment DF. Calculate the distance from point I to line AB according to a; infer that I is a fixed point, independent of the position of point M on segment AB.
d. Show that the line BE passes through I. When the point M moves on the line segment AB, on what fixed lines does the vertex C of the square BMEF move?
Question 3: A right angle xAy revolves around the vertex A of square ABCD. Side Ax intersects lines BC and CD respectively at points M, N and side Ay intersects lines BC, CD respectively at points P, Q.
a. Prove that the triangles NAP, MAQ are rightangled triangles.
b. Let E be the intersection of QM and PN; F and I are respectively the midpoints of the lines QM and PN. Prove that quadrilateral AFEU is a rectangle.
c. When the right angle xAy revolves around the vertex A, on which fixed line do the points F and I move?
3.2. Multiple choice exercises
Question 1: Which of the following signs is not enough to conclude that a quadrilateral is a square?
A. A rectangle with two equal adjacent sides is a square
B. A rectangle with two perpendicular diagonals is a square
C. A parallelogram with two equal diagonals is a square
D. A rectangle whose diagonal is the bisector of an angle is a square
Verse 2: Given a square with side 2a, the length of the diagonal of the square is
A. \(a\sqrt 2 \,\)
B. \(2a\sqrt 2 \)
C. \(\frac{{a\sqrt 2 }}{2}\)
D. \(\frac{{3a}}{{\sqrt 2 }}\)
Question 3: Given the diagonal of the square is \(4\sqrt 2 \) cm, what is the area of the square?
A. 32cm^{2}
B. 16cm^{2}
C. 8cm^{2}
D. 2cm^{2}
Question 4: A square has a perimeter of 28 cm, which of the following is the area of the square?
A. 49cm^{2}
B. 49cm^{2}
C. 14cm^{2}
D. 14cm^{2}
Question 5: Choose the wrong idea
A. A rhombus with one right angle is a square.
B. If a quadrilateral is both a rhombus and a rectangle, then it is a square.
C. A quadrilateral with four equal sides is a square.
D. A rectangle with two equal adjacent sides is a square.
4. Conclusion
Through this Square lesson, students need to complete some of the objectives given by the lesson, such as:

Identify squares.

Remember the properties and signs of squares.

Apply knowledge to solve some related problems.
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