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Solution for Exercise 5: The perpendicular bisector of a line segment (C8 Math 7 Horizon)
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Solution 1 page 70 Math textbook 7 Creative horizon volume 2 – CTST
Figure 10 illustrates a piece of paper with a drawing of the median xy line of line segment AB, where the image of point B is blurred. Explain how point B is determined.
Detailed instructions for solving Lesson 1
Solution method
– We find the intersection of the median and the line segment AB
– Then from that point find point B such that the distance from that point to A is equal to B and B, A and the intersection must be collinear, B does not coincide with A
Detailed explanation
Let the intersection of AB and xy be O .
\( \Rightarrow \) O is the midpoint of AB ( Do xy is the perpendicular bisector of AB)
\( \Rightarrow \) Measure the distance AO and from point O draw OB such that OA = OB and lies on the opposite side of point A from the line xy ( A, B, O align )
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Solution 2 page 70 Math textbook 7 Creative horizon volume 2 – CTST
Observe Figure 11, show that M is the midpoint of BC, AM is perpendicular to BC and AB = 10 cm, Calculate AC.
Detailed instructions for solving Lesson 2
Solution method
We prove that the two triangles AMC and AMB are congruent
Detailed explanation
Consider triangle AMB and triangle AMC with:
AM common edge
MB = MC (since M is the midpoint of BC)
\(\widehat {BMA} = \widehat {CMA} = {90^o}\)
\( \Rightarrow \) Triangle AMB = Triangle AMC (cgc)
\( \Rightarrow \)AB = AC = 10 cm (corresponding sides are equal)
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Solution 3 page 70 Math textbook 7 Creative horizon volume 2 – CTST
Observe Figure 12, show that AM is the perpendicular bisector of line segment BC and DB = DC = 8 cm. Prove that the three points A, M, D are collinear.
Detailed instructions for solving Lesson 3
Solution method
Prove that D belongs to the perpendicular bisector of BC \( \Rightarrow \) A, M, D are collinear
Detailed explanation
Consider triangle BCD with BD = CD (assumption )
\( \Rightarrow \) D belongs to the perpendicular bisector of BC because it is equidistant from the two ends of segment BC
Where AM is the perpendicular bisector of BC
\( \Rightarrow \) D belongs to the line AM
\( \Rightarrow \) A, M, D align
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Solution 4 page 70 Math textbook 7 Creative horizon volume 2 – CTST
Observe Figure 13, know AB = AC, DB = DC. Prove that M is the midpoint of BC.
Detailed instructions for solving Lesson 4
Solution method
– We prove that the two triangles ABD and ACD are congruent
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– Infer 2 triangles BAM and CAM are equal \( \Rightarrow \) BM = CM
Detailed explanation
Consider triangle ABD and triangle ACD with:
AB = AC (assumption )
BD = CD (assumption)
AD common edge
\( \Rightarrow \Delta ABD =\Delta ACD (ccc)\)
\( \Rightarrow \)\(\widehat {BAD} = \widehat {CAD}\)( 2 corresponding angles )
Consider triangle ABM and triangle ACM:
AB = AC (assumption )
AM common edge
\(\widehat {BAD} = \widehat {CAD}\)(proved above )
\(\Delta ABM=\Delta ACM (cgc)\)
\(\Rightarrow MC = MB\) ( 2 corresponding edges )
\( \Rightarrow \) M is the midpoint of BC
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Solve problems 5 pages 70 Math textbook 7 Creative horizon volume 2 – CTST
Let two points M and N lie on the perpendicular bisector d of the line segment EF. Prove that \(\Delta EMN=\Delta FMN\)
Detailed instructions for solving Lesson 5
Solution method
Prove that 2 triangles are congruent by case (ccc)
Detailed explanation
Since M belongs to the orthogonal EF, ME = MF (property of the point on the orthogonal )
Similarly \( \Rightarrow \) NE = NF (the property of the point on the orthogonal )
Consider two triangles MEN and MFN with :
MN is a common edge
ME = MF
NE = NF
\(\Rightarrow \Delta MEN = \Delta MFN (ccc)\)
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Solve problem 6 page 70 Math textbook 7 Creative horizon volume 2 – CTST
On the planning map of a residential area, there is a road d and two residential points A and B (Figure 14). Find on the roadside a location M to build a medical station such that the medical station is equidistant from two residential points.
Detailed instructions for solving Lesson 6
Solution method
– Let MA = MB \( \Rightarrow \) M belongs to the orthogonal AB
– Find M in d
Detailed explanation
Let M be equidistant from A, B if and only if M is on the perpendicular bisector of AB
Since M must belong to d \( \Rightarrow \) M is the intersection of the perpendicular bisector AB and the line d
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