adsense

**SBT Solution Lesson 7: Properties of the three medians of a triangle (C8 SBT Math 7 Horizon)**

===========

### Solution 1 page 60 SBT Math 7 Creative horizon episode 2 – CTST

Let ABC be a triangle with median AM and G as the centroid. Prove:

a) \({S_{AMB}} = {S_{AMC}}\)

b) \({S_{ABG}} = 2{S_{BMG}}\)

c) \({S_{GAB}} = {S_{GBC}} = {S_{GAC}}\)

**Detailed instructions for solving Lesson 1**

**Solution method**

Compare the altitudes and corresponding bases of the triangles

**Detailed explanation**

**a) Draw altitude AH of triangle ABC.**

Two triangles AMB and AMC have the same altitude AH and the same base side: BM = CM

Derive: \({S_{AMB}} = {S_{AMC}}\)(because \({S_{AMB}} = \frac{1}{2}.AH.BM{;^{}}{) S_{AMC}} = \frac{1}{2}.AN.CM\))

b) Draw altitude BK of triangle BGM.

Two triangles ABG and BMG have the same altitude BK and base side AG = 2MG.

Derive: \({S_{ABG}} = \frac{1}{2}.BK.AG = \frac{1}{2}.BK.2MG = 2.\frac{1}{2}.BK .MG = 2{S_{BMG}}\)

c) We have:

\({S_{ABG}} = \frac{2}{3}{S_{ABM}} = \frac{1}{3}{S_{ABC}}\)

Similarly: \({S_{ACG}} = \frac{2}{3}{S_{ACM}} = \frac{1}{3}{S_{ABC}}\)

Derive: \({S_{BCG}} = \frac{1}{3}{S_{ABC}}\)

So: \({S_{GAB}} = {S_{GBC}} = {S_{GAC}} = \frac{1}{3}{S_{ABC}}\)

–>

**— *******

### Solution 2 page 60 SBT Math 7 Creative horizon episode 2 – CTST

Let ABC be a triangle with median AM and bisector of angle A. Prove that triangle ABC is an isosceles triangle.

**Detailed instructions for solving Lesson 2**

**Solution method**

– Prove: \(\Delta AMH = \Delta AMK\) infer: MH = MK

– Prove: \(\widehat B = \widehat C\) deduce triangle ABC is isosceles

**Detailed explanation**

Draw altitude MH of triangle AMB and draw altitude MK of triangle AMC.

We have \(\Delta AMH = \Delta AMK\) (because two right triangles have a common hypotenuse AM, and an acute angle is equal)

Infer: MH = MK.

From that, we have: \(\Delta MBH = \Delta MCK\) (two right-angled triangles sharing hypotenuse Am and one equal right angle side: MH = MK)

Derive \(\widehat B = \widehat C\)

So triangle ABC is an isosceles triangle at A.

–>

**— *******

### Solution 3 page 60 SBT Math 7 Creative horizon episode 2 – CTST

Let ABC be a triangle with two medians AM and CN intersecting at G.

a) Knowing AM = 12 cm, calculate AG.

b) Knowing GN = 3 cm, calculate CN.

c) Find x knowing AG = 3x – 4, GM = x.

adsense

**Detailed instructions for solving Lesson 3**

**Solution method**

Using the property of three medians of a triangle

**Detailed explanation**

a) AM = 12 cm, so: \(AG = \frac{2}{3}AM = \frac{2}{3}.12 = 8cm\)

b) GN = 3cm inferred: CN = 3GN = 3.3= 9cm

c) AG = 2GM, infer: 3x – 4 = 2x infer x = 4

–>

**— *******

### Solution 4 page 60 SBT Math 7 Creative horizon episode 2 – CTST

Let ABC be a triangle with three medians AM, BN, CP concurrent at G. Prove: \(GA + GB + GC = \frac{2}{3}\left( {AM + BN + CP} \right) \)

**Detailed instructions for solving Lesson 4**

**Solution method**

Using property of three medians

**Detailed explanation**

We have G as the centroid of triangle ANC, so we have:

\(GA = \frac{2}{3}AM;GB = \frac{2}{3}BN;GC = \frac{2}{3}CP\)

Derive: \(GA + GB + GC = \frac{2}{3}\left( {AM + BN + CP} \right)\)

–>

**— *******

### Solve problems 5 pages 60 SBT Math 7 Creative horizon episode 2 – CTST

Let ABC be a triangle with two medians AM and BN intersecting at G. Draw AH perpendicular to BC at H. Show HB = HM. Prove:

a) \(\Delta ABH = \Delta AMH\)

b) \(AG = \frac{2}{3}AB\)

**Detailed instructions for solving Lesson 5**

**Solution method**

– Check the corresponding three sides of two triangles ABH and triangle AMH

– Using property of three medians

**Detailed explanation**

a) We have AH is the perpendicular bisector of segment BM, so AB = AM.

Consider two triangles ABH and AMH with:

Common AH edge

HB = HM

AB = AM

Derive: \(\Delta ABH = \Delta AMH(c – c – c)\)

b) G is the centroid of triangle ABC.

Derive: \(AG = \frac{2}{3}AM\)

According to sentence a we have: \(AB = AM\)

Derive: \(AG = \frac{2}{3}AB\)

–>

**— *******