Jump to heading🔗 Module 5-03 Parallel Lines

Jump to heading🔗 1.Angle Between a Straight Line and a parallel Line

• $\mathrm{\angle }1\text{and}\mathrm{\angle }4$ are corresponding angles, corresponding angles are equal.
• $\mathrm{\angle }2\text{and}\mathrm{\angle }4$ are alternate interior angles, alternate interior angles are equal.
• $\mathrm{\angle }3\text{and}\mathrm{\angle }4$ are adjacent interior angles, adjacent interior angles are supplementary.
  

Jump to heading🔗 2.Segments Intercepted by a Set of Parallel Lines Are Proportional

• Up-down ratio $\frac{a}{b}=\frac{c}{d}$.
• Left-right ratio $\frac{a}{c}=\frac{b}{d}$.
  

Jump to heading🔗 3.Focus 1

Solve angle

• Note that the angle formed by the combination of parallel lines and other special figures not only has the relationship between the angles of parallel lines, but also the relationship between the angle of special figures.
• Special figures similar Right triangle, Isosceles triangle.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{D}\mathbf{)}$
  According to the Solution, get $\mathrm{\angle }\alpha ={65}^{\circ }$, so choose $D$.

• Formula used

  $\begin{array}{l}\text{Angle relationship}\{\begin{array}{l}\mathrm{\angle }1=\mathrm{\angle }4\\ \mathrm{\angle }2=\mathrm{\angle }4\\ \mathrm{\angle }1+\mathrm{\angle }4={180}^{\circ }\end{array}\end{array}$

• Another solution $l_2$ elongation

  
  $\begin{array}{ll}\mathrm{\angle }4={180}^{\circ }-\mathrm{\angle }1& \text{Corresponding angles}\\ {180}^{\circ }-{105}^{\circ }={75}^{\circ }\\ \mathrm{\angle }\alpha +\mathrm{\angle }4=\mathrm{\angle }2& \text{Alternate interior angles}\\ \mathrm{\angle }\alpha =\mathrm{\angle }2-\mathrm{\angle }4={140}^{\circ }-{75}^{\circ }={65}^{\circ }\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{C}$
  According to the Solution, get $\mathrm{\angle }\alpha ={85}^{\circ }$, so choose $C$.

• Formula used

  $\begin{array}{l}\text{Angle relationship}\{\begin{array}{l}\mathrm{\angle }1=\mathrm{\angle }4\\ \mathrm{\angle }2=\mathrm{\angle }4\\ \mathrm{\angle }1+\mathrm{\angle }4={180}^{\circ }\end{array}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{C}\mathbf{)}$
  According to the Solution, get $\mathrm{\angle }M={50}^{\circ }$, so choose $C$.

• Formula used

  $\begin{array}{l}\text{Angle relationship}\{\begin{array}{l}\mathrm{\angle }1=\mathrm{\angle }4\\ \mathrm{\angle }2=\mathrm{\angle }4\\ \mathrm{\angle }1+\mathrm{\angle }4={180}^{\circ }\end{array}\\ \mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }& \text{The sum of the interior angles of a triangle is }{180}^{\circ }\end{array}$

Jump to heading🔗 4.Focus 2

Solve length

• Analysis based on the parallel line segments ratio formula.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{D}\mathbf{)}$
  According to the Solution, get $EF=6$, so choose $D$.

• Formula used

  $\begin{array}{ll}\frac{a}{c}=\frac{b}{d}& \text{Left-right ratio}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{C}\mathbf{)}$
  According to the Solution, get $AC=12.5$, so choose $C$.

• Formula used

  $\begin{array}{l}\text{Parallel line segments proportion}\{\begin{array}{l}\frac{a}{b}=\frac{c}{d}\\ \frac{a}{c}=\frac{b}{d}\end{array}\end{array}$

Jump to heading🔗 Module 6-02 Triangle

Jump to heading🔗 1.Sum of Interior Angles of a Triangle

$\mathrm{\angle }1+\mathrm{\angle }2+\mathrm{\angle }3=\pi$, the exterior angle of a triangle is equal to the sum of its two non-adjacent interior angles.

Jump to heading🔗 2.Relationship Between the Three Sides of a Triangle

• The sum of any two sides is greater than the third side, then $a+b>c$.
  $a+b>c,a+c>b,b+c>a$
• The difference between any two sides is less than the third side, then $|a-b|<c$.
  $|a-b|<c,|a-c|<b,|b-c|<a$

Jump to heading🔗 3.Focus 1

Solve angle

• Note: The angles formed when parallel lines are combined with other special figures not only have the relationship between the angles of parallel lines, but also the relationship between the angles of special figures.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{C}\mathbf{)}$
  According to the Solution, get $x={67.5}^{\circ }$, so choose $C$.

• Formula used

  $\begin{array}{l}\text{Angle relationship}\{\begin{array}{l}\mathrm{\angle }1=\mathrm{\angle }4\\ \mathrm{\angle }2=\mathrm{\angle }4\\ \mathrm{\angle }1+\mathrm{\angle }4={180}^{\circ }\end{array}\\ \mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }& \text{The sum of the interior angles of a triangle is }{180}^{\circ }\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{C}\mathbf{)}$
  According to the Solution, get $\mathrm{\angle }A=\frac{\pi }{10}$, so choose $C$.

• Formula used

  $\begin{array}{ll}A\perp B={90}^{\circ }& \text{Right-Angled triangle}\\ {90}^{\circ }=\frac{\pi }{2}& \text{Degree to radian conversion}\\ \text{Exterior angle of a triangle}=\mathrm{\angle }A+\mathrm{\angle }B\end{array}$

Jump to heading🔗 4.Focus 2

Trilateral relations

• According to the relationship between the three sides of a triangle, the requirements of a triangle can be analyzed, The sum of any two sides is greater than the third side, and the difference between any two sides is less third side, As long as one of the two sides is met, a triangle can be formed.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{A}\mathbf{)}$
  According to the Solution, $13$ types of triangles can be formed, so choose $A$.

• Formula used

  $\begin{array}{ll}\boxed{{\displaystyle \begin{array}{l}a+b>c,a+c>b,b+c>a\\ \text{Know the maximum side of c}\\ a+b>\underset{\mathrm{△}}{c}\Rightarrow (a+\underset{\mathrm{△}}{c}>b)\wedge (b+\underset{\mathrm{△}}{c}>a)\end{array}}}⟺\boxed{{\displaystyle \begin{array}{l}|a-b|<c,|a-c|<b,|b-c|<a\\ \text{Know the minimum side of c}\\ |a-b|<\underset{\mathrm{△}}{c}\Rightarrow \underset{\begin{array}{l}a-c<b\\ a<b+c\end{array}}{\underbrace{(|a-\underset{\mathrm{△}}{c}|<b)}}\wedge \underset{\begin{array}{l}b-c<a\\ b<a+c\end{array}}{\underbrace{(|b-\underset{\mathrm{△}}{c}|<a)}}\end{array}}}& \text{Trilateral relations}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{C}\mathbf{)}$
  According to the Solution, get the longest side length, which is $5$, so choose $C$.

• Formula used

  $\begin{array}{ll}\boxed{{\displaystyle \begin{array}{l}a+b>c,a+c>b,b+c>a\\ \text{Know the maximum side of c}\\ a+b>\underset{\mathrm{△}}{c}\Rightarrow (a+\underset{\mathrm{△}}{c}>b)\wedge (b+\underset{\mathrm{△}}{c}>a)\end{array}}}⟺\boxed{{\displaystyle \begin{array}{l}|a-b|<c,|a-c|<b,|b-c|<a\\ \text{Know the minimum side of c}\\ |a-b|<\underset{\mathrm{△}}{c}\Rightarrow \underset{\begin{array}{l}a-c<b\\ a<b+c\end{array}}{\underbrace{(|a-\underset{\mathrm{△}}{c}|<b)}}\wedge \underset{\begin{array}{l}b-c<a\\ b<a+c\end{array}}{\underbrace{(|b-\underset{\mathrm{△}}{c}|<a)}}\end{array}}}& \text{Trilateral relations}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $\frac{3}{2}<a<5$, so choose $B$.

• Formula used

  $\begin{array}{ll}\boxed{{\displaystyle \begin{array}{l}a+b>c,a+c>b,b+c>a\\ \text{Know the maximum side of c}\\ a+b>\underset{\mathrm{△}}{c}\Rightarrow (a+\underset{\mathrm{△}}{c}>b)\wedge (b+\underset{\mathrm{△}}{c}>a)\end{array}}}⟺\boxed{{\displaystyle \begin{array}{l}|a-b|<c,|a-c|<b,|b-c|<a\\ \text{Know the minimum side of c}\\ |a-b|<\underset{\mathrm{△}}{c}\Rightarrow \underset{\begin{array}{l}a-c<b\\ a<b+c\end{array}}{\underbrace{(|a-\underset{\mathrm{△}}{c}|<b)}}\wedge \underset{\begin{array}{l}b-c<a\\ b<a+c\end{array}}{\underbrace{(|b-\underset{\mathrm{△}}{c}|<a)}}\end{array}}}& \text{Trilateral relations}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $1<AD<4$, so choose $B$.

• Formula used

  $\begin{array}{ll}\boxed{{\displaystyle \begin{array}{l}a+b>c,a+c>b,b+c>a\\ \text{Know the maximum side of c}\\ a+b>\underset{\mathrm{△}}{c}\Rightarrow (a+\underset{\mathrm{△}}{c}>b)\wedge (b+\underset{\mathrm{△}}{c}>a)\end{array}}}⟺\boxed{{\displaystyle \begin{array}{l}|a-b|<c,|a-c|<b,|b-c|<a\\ \text{Know the minimum side of c}\\ |a-b|<\underset{\mathrm{△}}{c}\Rightarrow \underset{\begin{array}{l}a-c<b\\ a<b+c\end{array}}{\underbrace{(|a-\underset{\mathrm{△}}{c}|<b)}}\wedge \underset{\begin{array}{l}b-c<a\\ b<a+c\end{array}}{\underbrace{(|b-\underset{\mathrm{△}}{c}|<a)}}\end{array}}}& \text{Trilateral relations}\\ \frac{|AB-AC|}{2}<AD<\frac{AB+AC}{2}& \text{The formula for the median range of the third side in a triangle}\end{array}$

• For ranges, the limit solution can be used

  • $\mathrm{\angle }A\to 0^{\circ }$
    
    Note that D is the midpoint of BC.
    $AD\to 4$
  • $\mathrm{\angle }A\to {180}^{\circ }$
    
    $AD\to 1$

  ---

  $\boxed{{\displaystyle \begin{array}{l}AD\to 4\\ AD\to 1\end{array}}}\stackrel{\stackrel{\text{Intersection}}{}}{\to }1<AD<4$