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**Solution for Exercise 7: Properties of the three medians of a triangle (C8 Math 7 Horizons)**

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### Solution 1 page 75 Math textbook 7 Creative horizon volume 2 – CTST

Observe Figure 8.

Find the correct number to fill in the dot in the following equations:

\(\begin{array}{l}EG = …EM;\,\,\,GM = …EM;\,\,\,\,GM = …EG\\FG = …GN;\,\,\ ,\,\,FN = …GN;\,\,\,\,\,FN = …FG\end{array}\)

**Detailed instructions for solving Lesson 1**

**Solution method**

We rely on the theorem that three medians intersect at a point. That point is a distance from each vertex equal to \(\dfrac{2}{3}\) the length of the median passing through that vertex.

**Detailed explanation**

\(\begin{array}{l}EG = \dfrac{2}{3}EM;\,\,\,GM = \dfrac{1}{3}EM;\,\,\,\,GM = \dfrac{1}{2}EG\\FG = 2GN;\,\,\,\,\,FN = 3GN;\,\,\,\,\,\,\,FN = \dfrac{3} {2}FG\end{array}\)

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### Solve problem 2 page 75 Math textbook 7 Creative horizon volume 2 – CTST

Observe Figure 9

a) Knowing AM = 15 cm, calculate AG

b) Knowing GN = 6 cm, calculate CN

**Detailed instructions for solving Lesson 2**

**Solution method**

– We rely on the theorem of three medians intersecting at 1 point. That point is a distance from each vertex equal to \(\dfrac{2}{3}\)the length of the median passing through that vertex.

– We apply the ratios between the lines and their lengths

**Detailed explanation**

a) According to the problem, we have AM = 15 cm

Where CN and AM are the two medians of triangle ABC

AM intersects CN at G, so G is the centroid of triangle ABC

\( \Rightarrow AG = \dfrac{2}{3}AM\)(center of triangle theorem)

\( \Rightarrow AG = \dfrac{2}{3}\,15cm = 10cm\)

b) Since G is the centroid of triangle ABC

\( \Rightarrow CG = \dfrac{2}{3}CN\)(according to the property of the median passing through the centroid)

And \(CG + GN = CN\) so we have \(GN = CN – CG = CN – \dfrac{2}{3}CN = \dfrac{1}{3}CN\)

Under the assumption GN = 6cm we have

\( \Rightarrow CN = 3GN = 3.6cm = 18cm\)

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### Solution 3 page 75 Math textbook 7 Creative horizon volume 2 – CTST

Let ABC be a triangle. The two medians AM and CN intersect at G. On the opposite ray of the ray AM take the point E such that ME = MG.

a) Prove that BG is parallel to EC.

b) Let I be the midpoint of BE, AI intersects BG at F. Prove that AF = 2FI

**Detailed instructions for solving Lesson 3**

**Solution method**

– We rely on the theorem of three medians intersecting at 1 point. That point is a distance from each vertex equal to \(\dfrac{2}{3}\)the length of the median passing through that vertex.

– Sentence a will prove that 2 staggered interior angles are congruent through congruent triangles

– Sentence b will prove that F is the centroid of triangle ABE

**Detailed explanation**

a) Consider triangle BGM and triangle CEM:

\(\widehat {GMB} = \widehat {EMC}\)(2 opposite angles)

GM = ME (because G is symmetric about E through M)

MB = MC (since M is mid point of BC)

\( \Rightarrow \Delta BGM = \Delta CEM(c – g – c)\)

\( \Rightarrow \widehat {GBM} = \widehat {MCE}\)(2 corresponding angles are equal)

The two upper corners are in staggered positions, so BG⫽CE

b) Since I is the midpoint of BE, AI will be the median of triangle ABE

And BG is also the median of triangle ABE since G is the midpoint of AE

Since BG intersects AI at F, F will be the centroid of triangle ABE

\(\, \Rightarrow AF = \dfrac{2}{3}AI\)(triangle centroid theorem)

Which AI = AF + FI \( \Rightarrow \) FI = AI – AF

\( \Rightarrow FI = AI – \dfrac{2}{3}AI = \dfrac{1}{3}AI\)

\( \Rightarrow 2FI = AF = \dfrac{2}{3}AI\)

\( \Rightarrow \) AF = 2 FI

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### Solve problems 4 pages 75 Math textbook 7 Creative horizon volume 2 – CTST

Let ABC be an isosceles triangle at A with BM and CN as the two medians.

a) Prove that BM = CN

b) Let I be the intersection of BM and CN, the line AI intersects BC at H. Prove that H is the midpoint of BC.

**Detailed instructions for solving Lesson 2**

**Solution method**

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– We prove that two triangles are congruent so that we can prove that the two segments are congruent

– We prove that I is the centroid of triangle ABC and prove that AH is the median of triangle ABC and H is the midpoint of BC.

**Detailed explanation**

a) Because triangle ABC is isosceles at A by assumption. BM and CN are 2 medians, so M and N are the midpoints of AC and AB.

Because AB = AC (isosceles triangle property)

\( \Rightarrow \dfrac{{AB}}{2} = \dfrac{{AC}}{2} = AN = AM\)

Considering triangle AMB and triangle ANC, we have:

AM = AN (cmt)

AB = AC

Common angle A

\( \Rightarrow \Delta AMB =\Delta ANC\)

\( \Rightarrow BM = CN\) ( 2 corresponding edges )

b) Since BM and CN are medians

Where I is the intersection of BM and CN

\( \Rightarrow \) I is the centroid of triangle ABC

\( \Rightarrow \) AI is the median of triangle ABC or AH is the median of triangle ABC

\( \Rightarrow \) H is the midpoint of BC

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### Solve problems 5 pages 76 Math textbook 7 Creative horizon volume 2 – CTST

Let ABC be a triangle with median BM equal to median CN. Prove that triangle ABC is isosceles.

**Detailed instructions for solving Lesson 5**

**Solution method**

We prove AB = AC by proving that the two triangles are congruent

**Detailed explanation**

Let D be the intersection of CN and BM .

\( \Rightarrow \) D is the centroid of triangle ABC

\( \Rightarrow CD = \dfrac{2}{3}CN = BD = \dfrac{2}{3}BM\) ( do BM = CN )

\( \Rightarrow \) triangle DBC is isosceles at D since BD = CD

\( \Rightarrow \) \(\widehat {DBC} = \widehat {DCB}\)(2 base angles in an isosceles triangle) (1)

Consider \(\Delta NDB\) and \(\Delta MDC\) with :

BD = CD

\(\widehat {NDB} = \widehat {MDC}\) (2 opposite angles)

ND = DM (due to the same \( = \dfrac{1}{3}CN = \dfrac{1}{3}BM\) (property of the orthocenter passing through the centroid of the triangle ))

\( \Rightarrow \Delta NDB=\Delta MDC\) (cgc)

\( \Rightarrow \,\widehat {NBD} = \widehat {MCD}\)(2 corresponding angles) (2)

From (1) and (2) \( \Rightarrow \widehat {ABC} = \widehat {ACB}\) do \(\widehat {ABC} = \widehat {NBD} + \widehat {DBC}\) and \( \widehat {ACB} = \widehat {MCD} + \widehat {DCB}\)

\( \Rightarrow \Delta ABC\) is equal at A (due to 2 equal angles)

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### Solve problem 6 page 76 Math textbook 7 Creative horizon volume 2 – CTST

Let ABC be an isosceles triangle at A with BD and CE being the two medians intersecting at F (Figure 10). Knowing BE = 9 cm, calculate the length of line segment DF.

**Detailed instructions for solving Lesson 6**

**Solution method**

– We prove that F is the centroid of triangle ABC

– Then prove CD = BE

– Applying the theorem about the center of gravity of the triangle, we calculate the segments DF, EF

**Detailed explanation**

Since BE and CD are the two medians of triangle ABC

SO E, D are medians of AB and AC . respectively

\( \Rightarrow AD = AE = \dfrac{1}{2}AB = \dfrac{1}{2}AC\)

Consider triangle ADC and triangle AEB:

AD = AE

Common angle A

AB = AC (triangle ABC is isosceles at A by assumption)

\( \Rightarrow \Delta ADC = \Delta AEB(c – g – c)\)

\( \Rightarrow BE = CD\)(corresponding edge)

Since F is the intersection of two medians, F is the centroid of triangle ABC

\( \Rightarrow CF = BF = \dfrac{2}{3}BE = \dfrac{2}{3}CD\) ( theorem about median passing through centroid of triangle )

\( \Rightarrow \dfrac{1}{3}BE = \dfrac{1}{3}CD \Rightarrow DF = FE = \dfrac{1}{3}.9cm = 3cm\)

\( \Rightarrow \) DF = 3 cm

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