【PEP】Junior High School Math, Grade 8, Semester 2: The Mystery of Overlapping Strips

Imagine parallel light rays from physics passing through holes in a cardboard sheet, creating a shadow pattern on a table, or simply cutting two transparent strips with parallel edges and randomly overlapping them. No matter how you rotate the angle of these two strips, the overlapping dark region under light will always form a perfect geometric shape—parallelogram.

The Essence and Decomposition of Parallelograms

In geometry, 'parallel' represents an order of never intersecting lines. When we combine two pairs of mutually parallel line segments, we define this fascinating polygon:A quadrilateral with two pairs of opposite sides parallel is called a parallelogram(denoted as $\square ABCD$).

To unlock the mystery of the parallelogram, mathematicians employ a brilliant strategy of dimension reduction:connecting the diagonals. One diagonal instantly cuts the unknown quadrilateral into two triangles we already know!

Step 1: Introduce the diagonal to build a bridge

As shown in Figure 18.1-3, connect diagonal $AC$ in $\square ABCD$.

Using the 'alternate interior angles' magic of parallel lines:
$\because AD \parallel BC$ and $AB \parallel CD$
$\therefore \angle 1 = \angle 2$, and $\angle 3 = \angle 4$.

Step 2: Victory of Congruent Triangles

At this point, $AC$ is thecommon side.

According to the Angle-Side-Angle (ASA) theorem, $\therefore \triangle ABC \cong \triangle CDA$.
Once congruent, corresponding elements are completely equal:
$\therefore AD=CB$, $AB=CD$, and $\angle B=\angle D$.

Distance and Height: The Eternal Harmony of Parallel Lines

Why is the height always the same for the same base, no matter how the parallelogram is tilted? This leads us to another core concept:the distance between parallel lines. The perpendicular segment from any point on one line to the other line is called the distance between two parallel lines. Just like the sleepers between railway tracks, their length remains constant.

🎯 Core Principles and Criteria for Identification

Once you master the technique of decomposing into congruent triangles, you can easily derive all properties and criteria for identification!

Properties Theorem:Opposite sides of a parallelogram are equal; opposite angles are equal; diagonals bisect each other.
Identification Theorems (Reverse Reasoning):A quadrilateral with two pairs of equal opposite sides is a parallelogram; a quadrilateral with two pairs of equal opposite angles is a parallelogram; a quadrilateral with bisecting diagonals is a parallelogram; a quadrilateral with one pair of parallel and equal opposite sides is a parallelogram.

QUESTION 1

In parallelogram $\square ABCD$, given adjacent sides $AB=5$ and $BC=3$, find the perimeter of the parallelogram.

QUESTION 2

In parallelogram $\square ABCD$, given interior angle $\angle A=38^{\circ}$, what are the measures of $\angle B$ and $\angle C$?

$\angle B=142^{\circ}$, $\angle C=38^{\circ}$

$\angle B=38^{\circ}$, $\angle C=142^{\circ}$

$\angle B=142^{\circ}$, $\angle C=142^{\circ}$

$\angle B=52^{\circ}$, $\angle C=38^{\circ}$

QUESTION 3

After walking 80 meters east, Xiao Ming walks another 60 meters in a different direction, then 100 meters in a third direction and returns exactly to the starting point. After walking 80 meters east, in which direction did he walk the second time?

South or North

Northeast or Southeast

East or West

Northwest or Southwest

QUESTION 4

To ensure the two straight railway tracks are absolutely parallel, workers only need to make sure all sleepers placed between them have equal length. What geometric principle does this embody?

Parallel line segments between two parallel lines are equal in length, and the distance between parallel lines is constant everywhere

Corresponding angles of congruent triangles are equal

A quadrilateral with bisecting diagonals is a parallelogram

The midline of a triangle is parallel to the third side and half its length

QUESTION 5

As shown in the figure, $\square OABC$ is in the coordinate plane, with vertices $O, A, C$ at coordinates $(0,0), (a,0), (b,c)$. Using the properties of a parallelogram, what should be the coordinates of vertex $B$?

$(a+b, c)$

$(a, c)$

$(b-a, c)$

$(a+c, b)$

Practical Exercise: In-depth Exploration of Parallelograms

Solving complex problems using congruent triangles and identification theorems

[Hands-on Exploration] Cut two strips with parallel opposite edges, randomly overlap them, and the overlapping area forms a quadrilateral $ABCD$. When you rotate one strip, what is the relationship between the lengths of segments $AD$ and $BC$? Why?

Detailed Explanation:
Relationship:Segment $AD = BC$.
Reason:When the two strips are overlapped crosswise, because the opposite edges of the strips are parallel, the quadrilateral $ABCD$ formed by the overlapping region necessarily has $AB \parallel CD$ and $AD \parallel BC$. According to the definition 'a quadrilateral with two pairs of opposite sides parallel is a parallelogram', quadrilateral $ABCD$ is a parallelogram. Then, according to the property theorem 'opposite sides of a parallelogram are equal', we conclude $AD = BC$. This theorem holds true regardless of the rotation angle.

[Perimeter and Diagonals] In $\square ABCD$, diagonals $AC$ and $BD$ intersect at point $O$. Given $BC=10$, $AC=8$, $BD=14$.
(1) What is the perimeter of $\triangle AOD$?
(2) Which has a longer perimeter, $\triangle ABC$ or $\triangle DBC$? By how much?

Detailed Explanation:
(1) Find the perimeter of $\triangle AOD$:
Based on the properties of parallelograms 'diagonals bisect each other' and 'opposite sides are equal', we know:
$AO = \frac{1}{2}AC = 4$,
$DO = \frac{1}{2}BD = 7$,
$AD = BC = 10$.
Therefore, the perimeter of $\triangle AOD$ = $AD + AO + DO = 10 + 4 + 7 = 21$.

(2) Compare the perimeters of $\triangle ABC$ and $\triangle DBC$:
$\triangle ABC$'s perimeter = $AB + BC + AC = AB + 10 + 8 = AB + 18$.
$\triangle DBC$'s perimeter = $CD + BC + BD$. Since $CD = AB$, the perimeter is $AB + 10 + 14 = AB + 24$.
Upon comparison, $\triangle DBC$ has a longer perimeter than $\triangle ABC$, by $24 - 18 = 6$.

[Proof Challenge] As shown in the figure, diagonals $AC, BD$ of $\square ABCD$ intersect at point $O$, and line $EF$ passes through point $O$, intersecting $AB, CD$ at points $E, F$ respectively. Prove that $OE=OF$.

Proof Process:
In $\square ABCD$, opposite sides $AB \parallel CD$, and diagonals bisect each other, so $OA = OC$.
$\because AB \parallel CD$
$\therefore \angle OAE = \angle OCF$ (when two lines are parallel, alternate interior angles are equal).
In $\triangle AOE$ and $\triangle COF$:
$\begin{cases} \angle OAE = \angle OCF \\ OA = OC \\ \angle AOE = \angle COF \text{ (vertical angles are equal)} \end{cases}$
$\therefore \triangle AOE \cong \triangle COF$ (ASA Theorem).
$\therefore OE = OF$ (corresponding sides of congruent triangles are equal). Proof completed.
*Insight: This conclusion shows that any line passing through the center of a parallelogram not only bisects its area but also has its segment intercepted by opposite sides bisected at the center.

[Multi-Shape Interaction] As shown in the figure, there are two connected quadrilaterals $ABCD$ and $CDEF$, with $AB=DC=EF$, $AD=BC$, and $DE=CF$. What parallel line segments are present in the diagram?

Detailed Explanation:
There are three sets of parallel line segments in the diagram:$AB \parallel DC \parallel EF$,$AD \parallel BC$, and $DE \parallel CF$.
Reasoning Process:
1. In quadrilateral $ABCD$, since $AB=DC$ and $AD=BC$. According to the identification theorem 'a quadrilateral with two pairs of equal opposite sides is a parallelogram', quadrilateral $ABCD$ is a parallelogram. Therefore, $AB \parallel DC$ and $AD \parallel BC$.
2. Similarly, in quadrilateral $CDEF$, since $DC=EF$ (from $AB=DC=EF$) and $DE=CF$. Quadrilateral $CDEF$ is also a parallelogram. Therefore, $DC \parallel EF$ and $DE \parallel CF$.
3. Finally, since $AB \parallel DC$ and $DC \parallel EF$, by the transitive property of parallel lines, $AB \parallel EF$ must hold. Thus, the three segments $AB, DC, EF$ are mutually parallel.

[Algebra and Geometry Integration] If quadrilateral $ABCD$ is a parallelogram, with one side $AB=6$, and $AB$'s length is exactly $\frac{3}{16}$ of the perimeter of $\square ABCD$, what is the length of the adjacent side $BC$?

Detailed Explanation:
1. First, calculate the total perimeter based on $AB$'s length ratio. Perimeter $L = 6 \div \frac{3}{16} = 6 \times \frac{16}{3} = 32$.
2. According to the property that opposite sides of a parallelogram are equal, perimeter $= 2 \times (AB + BC)$.
3. Substitute known values into the equation: $32 = 2 \times (6 + BC)$.
4. Solving gives $6 + BC = 16$, so $BC = 10$.

[Conceptual Analysis] Is the quadrilateral satisfying the following conditions a square?
(Supplementary knowledge: If it's a right triangle, the three sides satisfy $a^2=c^2-b^2$. How does a quadrilateral evolve further?)
If the three side lengths $a, b, c$ of a triangle satisfy $a^2 = c^2 - b^2$, is the triangle a right triangle? Why?

Detailed Explanation:
Yes, it is a right triangle!
By simple algebraic rearrangement of the equation $a^2 = c^2 - b^2$, we get:$a^2 + b^2 = c^2$.
According tothe converse of the Pythagorean theorem, if the three side lengths of a triangle satisfy that the sum of the squares of two sides equals the square of the third side, then the triangle must be a right triangle (with $c$ as the hypotenuse).
*Extended Thinking:If a parallelogram is divided by its diagonals into two right triangles (i.e., the internal angles of the parallelogram are $90^{\circ}$), it evolves into a rectangle; if the adjacent sides are also equal, it ultimately becomes a square. The fundamental concepts of right angles and parallelism are the foundation for exploring all special quadrilaterals!