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SBT Solution Lesson 5: The perpendicular bisector of a line segment (C8 SBT Math 7 Horizon)
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Solution 1 page 55 SBT Math 7 Creative horizon episode 2 – CTST
Given three isosceles triangles MAB, NAB, PAB with common base AB. Prove that the three points M, N, and P are collinear.
Detailed instructions for solving Lesson 1
Solution method
Prove that the three points M, N, and P all belong to the median line of AB.
Detailed explanation
We have isosceles triangles MAB, NAB, PAB with the same base AB, infer MA = MB, NA = NB,
PA = PB. Points M, N, and P belong to the perpendicular bisector of AB, so they are collinear.
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Solution 2 page 55 SBT Math 7 Creative horizon episode 2 – CTST
Let angle xOy equal \({45^o}\) and point M in the angle xOy. Draw a point N such that Ox is the perpendicular bisector of MN, draw a point P such that Oy is the perpendicular bisector of MP.
a) Prove ON = OP.
b) Calculate the measure of angle NOP
Detailed instructions for solving Lesson 2
Solution method
Use the property of the median line to prove ON = OP
Detailed explanation
a) We have Ox as the perpendicular bisector of MN, so OM = ON
We have Oy as the median of MP, so OM = OP.
So ON = OP.
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b) Let H and K be the midpoints of MN and MP, respectively, so HM = HN ; KM = KP
We have: \(\Delta OHM = \Delta OHN\) (because OH is common, OM = ON, HM = HN)
\(\Delta OKM = \Delta OKP\) (because of general OK, OM = OP, KM = KP)
Derive: \(\widehat {NOP} = \widehat {NOM} + \widehat {MOP} = 2\left( {\widehat {xOM} + \widehat {MOy}} \right) = 2\widehat {xOy} = {2.45^o} = {90^o}\)
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Solve problem 3 page 55 SBT Math 7 Creative horizon episode 2 – CTST
Given two points A, B are the positions of the two factories on the same side of the river as the straight line a. Draw point C such that a is the perpendicular bisector of AC. Take an arbitrary point M on a.
a) Prove MA + MB \( \ge \) BC
b) Find the location of the location \({M_o}\) on the river bank to build a pumping station so that the total length of the pipeline from the pumping station to the two plants is the shortest.
Detailed instructions for solving Lesson 3
Solution method
Using with arbitrary points M, B, C we always have: MB + MC \( \ge \) BC to deduce the proof in the problem and find the position \({M_o}\)
Detailed explanation
a) We have a point M on the perpendicular bisector of AC, so MA = MC.
With three arbitrary points M, B, C we always have:
\(MB + MC \ge BC\)
So \(MA + MB \ge BC\)
b) We have: \(MA + MB \ge BC\), deduce the shortest MA + MB when B < C, M is collinear.
So the point \({M_o}\), to find is the intersection of line BC and line a.
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