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**Problem Solving Lesson 34 Convergence of three medians and three bisectors in a triangle (Chapter 9 Math 7 Connection)**

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### Solution 9.20 page 76 Math textbook 7 Connecting knowledge volume 2 – KNTT

Given triangle ABC with two medians BN, CP and centroid G. Find the appropriate number in the question mark to get the equality:

BG = ? BN, CG = ? CP;

BG = ? GN, CG = ? GP.

Detailed explanation guide

**Solution method**

+) Use the theorem about the concurrency of the three medians of the triangle.

+) Rule for adding straight lines.

**Detailed explanation**

Since G is the centroid of \(\Delta ABC\) \(BG = \dfrac{2}{3}BN,CG = \dfrac{2}{3}CP\)

We have: GN = BN – BG = BN – \(\dfrac{2}{3}\)BN = \(\dfrac{1}{3}\)BN; GP = CP – CG = CP – \(\dfrac{2}{3}\)CP = \(\dfrac{1}{3}\)CP

Therefore, BN = 3. GN ; CP = 3. GP

Thus, \(BG = \dfrac{2}{3}BN = \dfrac{2}{3}.3.GN = 2GN;CG = \dfrac{2}{3}CP = \dfrac{2}{ 3}.3.GP = 2GP\)

So \(BG = \dfrac{2}{3}BN,CG = \dfrac{2}{3}CP\);

BG = 2GN; CG = 2GP.

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### Solve problem 9.21 page 76 Math 7 Textbook Connecting knowledge volume 2 – KNTT

Prove that:

a) In an isosceles triangle, the two medians corresponding to the two sides are two equal line segments.

b) Conversely, if a triangle has two equal medians, then the triangle is isosceles.

Detailed instructions for solving problem 9.21

**Solution method**

Considering similar triangles, deduce the pair of corresponding sides that are congruent.

**Detailed explanation**

Let BM and CN be the medians of \(\Delta ABC\)

\( \Rightarrow \)MA = MC = \(\dfrac{1}{2}\)AC; NA = NB = \(\dfrac{1}{2}\)AB

Since \(\Delta ABC\) is balanced at A, AB = AC (property)

Therefore, AM = MC = NA = NB

Considering \(\Delta \)ANC and \(\Delta \)AMB, we have:

AN = AM

\(\widehat A\) common

AC = AB

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

\( \Rightarrow \) NC = MB ( 2 corresponding edges)

So the two medians corresponding to the two sides of an isosceles triangle are two equal line segments.

Since \(∆ABC\) has two medians \(BM\) and \(CN\) intersect at \(G\)

\(\Rightarrow \) \(G\) is the centroid of the triangle \(ABC\).

\(\Rightarrow GB = \dfrac{2}{3}BM\); \(GC = \dfrac{2}{3}CN\) (property of medians in triangles)

Which \(BM = CN\) (assumption) should \(GB = GC.\)

The triangle \(GBC\) has \(GB = GC\) so \(∆GBC\) is isosceles at \(G\).

\(\Rightarrow \) \(\widehat{GCB} = \widehat{GBC}\) (Property of isosceles triangle).

Consider \(∆BCN\) and \(∆CBM\) to have:

+) \(BC\) is a common edge

+) \(CN = BM\) (assumption)

+) \(\widehat{GCB} = \widehat{GBC}\) (proved above)

Derive \(∆BCN = ∆CBM\) (cgc)

\(\Rightarrow \) \(\widehat{NBC} = \widehat{MCB}\) (the two corresponding angles).

\(\Rightarrow ∆ABC\) isosceles at \(A\) (a triangle with two equal angles is an isosceles triangle)

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### Solve problem 9.22, page 76 Math 7 Textbook Connecting knowledge volume 2 – KNTT

Let the angle xOy be different from the flat angle. Use a compass to construct a circle with center O cutting Ox at A and intersecting Oy at B. Then construct two circles with center A and center B with equal radius so that they intersect at M and lie in the angle xOy. Prove that ray OM is the bisector of angle xOy.

Detailed instructions for solving problem 9.22

**Solution method**

Considering similar triangles, deduce pairs of corresponding angles that are congruent.

**Detailed explanation**

We have: AM = radius of the circle with center A

BM = radius of the circle with center B

These two circles have the same radius

Therefore, AM = BM

Consider \(\Delta \)OAM and \(\Delta \)ONM have:

OA = OB( = radius of circle center O)

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MA = MB

General OM

\( \Rightarrow \) \(\Delta \)OAM and \(\Delta \)ONM ( ccc)

\( \Rightarrow \) \(\widehat {AOM} = \widehat {BOM}\) ( 2 corresponding angles)

Where OM lies between two rays OA and OB

\( \Rightarrow \) OM is the bisector of angle AOB.

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### Solve problem 9.23 page 76 Math 7 Textbook Connecting knowledge volume 2 – KNTT

The symbol I is the point of convergence of the three bisectors of the triangle ABC. Calculate the angle BIC when the angle BAC is 120\(^\circ \).

Detailed instructions for solving problem 9.23

**Solution method**

Apply the bisector of an angle and the sum of the 3 angles in a triangle is 180 degrees.

**Detailed explanation**

Since BI is the bisector of angle ABC, \(\widehat {{B_1}} = \widehat {{B_2}} = \dfrac{1}{2}.\widehat {ABC}\)

Since CI is the bisector of angle ACB, \(\widehat {{C_1}} = \widehat {{C_2}} = \dfrac{1}{2}.\widehat {ACB}\)

Applying the triangle sum theorem in triangle ABC, we have:

\(\begin{array}{l}\widehat {BAC} + \widehat {ABC} + \widehat {ACB} = 180^\circ \\ \Rightarrow \widehat {ABC} + \widehat {ACB} = 180^ \circ – \widehat {BAC} = 180^\circ – 120^\circ = 60^\circ \\ \Rightarrow \widehat {{B_2}} + \widehat {{C_2}} = \dfrac{1}{2 }.\left( {\widehat {ABC} + \widehat {ACB}} \right) = \dfrac{1}{2}.60^\circ = 30^\circ \end{array}\)

Applying the theorem of the sum of three angles in triangle BIC, we have:

\(\begin{array}{l}\widehat {BIC} + \widehat {{B_2}} + \widehat {{C_2}} = 180^\circ \\ \Rightarrow \widehat {BIC} = 180^\circ – \left( {\widehat {{B_2}} + \widehat {{C_2}}} \right) = 180^\circ – 30^\circ = 150^\circ \end{array}\)

So \(\widehat {BIC} = 150^\circ \)

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### Solve problem 9.24 page 76 Math textbook 7 Connecting knowledge volume 2 – KNTT

Let BE and CF be the bisectors of an isosceles triangle ABC at A. Prove BE = CF.

Detailed instructions for solving problem 9.24

**Solution method**

Using the property of isosceles triangles, consider two congruent triangles and then indicate the two corresponding sides are congruent.

**Detailed explanation**

Since triangle ABC is isosceles at A, AB = AC; \(\widehat {ABC} = \widehat {ACB}\) (property)

Since BE is the bisector of angle ABC, \(\widehat {{B_1}} = \widehat {{B_2}} = \dfrac{1}{2}.\widehat {ABC}\)

Since CF is the bisector of angle ACB, \(\widehat {{C_1}} = \widehat {{C_2}} = \dfrac{1}{2}.\widehat {ACB}\)

Therefore, \(\widehat {{B_1}} = \widehat {{C_1}}\)

Considering \(\Delta ABE\) and \(\Delta ACF\), we have:

\(\widehat A\) common

AB = AC

\(\widehat {{B_1}} = \widehat {{C_1}}\)

\( \Rightarrow \Delta ABE = \Delta ACF\left( {gcg} \right)\)

\( \Rightarrow \)BE = CF ( 2 corresponding edges)

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### Solve problem 9.25 on page 76 Math 7 Textbook Connecting knowledge volume 2 – KNTT

In triangle ABC, the bisectors of angles B and C intersect at D. Draw DP perpendicular to BC, DQ perpendicular to CA, DR perpendicular to AB.

a) Explain why DP = DR.

b) Explain why DP = DQ.

c) From sentences a and b, deduce DR = DQ. Why does D lie on the bisector of angle A? (This is one way of proving theorem 2).

Detailed instructions for solving problem 9.25

**Solution method**

Using the property of the bisector of an angle, consider 2 congruent triangles, infer the corresponding sides are congruent.

**Detailed explanation**

a) Since BD is the bisector of angle ABC, \(\widehat {{B_1}} = \widehat {{B_2}} = \dfrac{1}{2}.\widehat {ABC}\)

Since CD is the bisector of angle ACB, \(\widehat {{C_1}} = \widehat {{C_2}} = \dfrac{1}{2}.\widehat {ACB}\)

Considering \(\Delta BDP\) square at P and \(\Delta BDR\) square at R, we have:

\(\widehat {{B_2}} = \widehat {{B_1}}\)

General BD

\( \Rightarrow \Delta BDP = \Delta BDR\) ( hypotenuse – acute angle)

\( \Rightarrow \) DP = DR ( 2 corresponding edges) (1)

b) Considering \(\Delta CDP\) square at P and \(\Delta CDQ\) square at Q, we have:

\(\widehat {{C_2}} = \widehat {{C_1}}\)

General CD

\( \Rightarrow \Delta CDP = \Delta CDQ\) ( hypotenuse – acute angle)

\( \Rightarrow \) DP = DQ ( 2 corresponding edges) (2)

c) From (1) and (2), we get: DR = DQ (same as DP).

D lies on the bisector of angle A since D is equidistant from AB and AC.

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