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**Solution of Exercise 3: Isosceles triangle (C8 Math 7 Horizon)**

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### Solve problem 1 page 62 Math textbook 7 Creative horizon volume 2 – CTST

Find the isosceles and equilateral triangles in each of the following figures (Figure 13). Explain.

**Detailed instructions for solving Lesson 1**

**Solution method**

Based on the sides and the measure of the base angles of each triangle

**Detailed explanation**

a) Triangle ABM is an equilateral triangle because it has 3 equal sides

Triangle AMC is isosceles at M since AM = MC

b) Triangle EDG is an equilateral triangle because it has 3 equal sides

Triangle EHF is isosceles at E since EH = EF

Triangle EDH is isosceles at D since DH = DE

c) The triangle EGF is isosceles at G since GE = GF

Triangle IHG is equilateral because it is an isosceles triangle with 1 angle = 60°

Triangle EHG is isosceles at E since EG = EH

d) triangle MBC is not isosceles and irregular because the 3 angles have different measures

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### Solve problem 2, page 62, Math 7 Textbook, Creative horizon, volume 2 – CTST

Given Figure 14, know ED = EF and EI is the bisector of \(\widehat {DEF}\)

Prove that:

a) \(\Delta EID = \Delta EIF\)

b) Isosceles triangle DIF

**Detailed instructions for solving Lesson 2**

**Solution method**

– We use the property cgc to prove statement a

– From sentence a, we infer ID = FI and prove that triangle DIF is isosceles

**Detailed explanation**

a) Consider triangle EID and triangle EIF with :

General IE

ED = EF

\(\widehat {IED} = \widehat {IEF}\)( EI is the bisector of \(\widehat {DEF}\))

\( \Rightarrow \Delta EID = \Delta EIF(c – g – c)\)

b) Since \(\Delta EID = \Delta EIF\) so ID = IF ( 2 corresponding edges )

Hence triangle DIF is isosceles at I (according to the definition of an isosceles triangle).

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### Solution 3 page 63 Math textbook 7 Creative horizon volume 2 – CTST

Let ABC be an isosceles triangle at A with \(\widehat A = {56^o}\)(Figure 15)

a) Calculate\(\widehat B\), \(\widehat C\)

b) Let M, N be the midpoints of AB and AC respectively. Prove that triangle AMN is isosceles.

c) Prove that MN // BC

**Detailed instructions for solving Lesson 3**

**Solution method**

a) Use the summation theorem of 3 angles in a triangle and property of the 2 base angles of an isosceles triangle

b) Prove AM = AN

c) Use isotopic angle property

**Detailed explanation**

a) According to the problem, we have an isosceles triangle ABC at A and \(\widehat A = {56^o}\)

Where \( \Rightarrow \widehat A + \widehat B + \widehat C = {180^o}\)

\( \Rightarrow \widehat B = \widehat C = ({180^o} – {56^o}):2 = {62^o}\)

b) Since triangle ABC is isosceles at A, AB = AC (definition of isosceles triangle)

Where M, N are midpoints of AB, AC

So AM = AN

Consider triangle AMN with AM = AN, so AMN is an isosceles triangle at A

\( \Rightarrow \widehat M = \widehat N = ({180^o} – {56^o}):2 = {62^o}\)

c) Because \(\widehat {AMN}=\widehat {ABC}\) (same as 62°)

Since they are in isotopic positions, MN⫽BC

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### Solution 4 page 63 Math textbook 7 Creative horizon volume 2 – CTST

Let ABC be an isosceles triangle at A (Figure 16). The bisector of angle B intersects AC at F, the bisector of angle C meets AB at E.

a) Prove that \(\widehat {ABF} = \widehat {ACE}\)

b) Prove that triangle AEF is isosceles

c) Let I be the intersection of BF and CE. Prove that triangle IBC and triangle IEF are isosceles triangles

**Detailed instructions for solving Lesson 4**

**Solution method**

a) Use properties of isosceles triangle and bisector

b) From sentence a infer AE = AF

adsense

c) Triangle IEF proves isosceles by proving that the two sides are congruent

Prove that IBC is isosceles because the base angles are congruent

**Detailed explanation**

a) Since triangle ABC is isosceles at A

\( \Rightarrow \widehat B = \widehat C \Rightarrow \dfrac{1}{2}\widehat B = \dfrac{1}{2}\widehat C \Rightarrow \widehat {ABF} = \widehat {ACE}\ )

b) Consider \(\Delta ECA\) and \(\Delta FBA\) have:

\(\widehat{A}\) general

AB = AC

\(\widehat {ABF} = \widehat {ACE}\)

\( \Rightarrow \)\(\Delta ECA\)= \(\Delta FBA\)( g – c – g )

\( \Rightarrow AE = AF and EC = BF\) (2 sides respectively)

\( \Rightarrow \Delta AEF\) weighed at A

c) Consider triangle IBC with :

\(\widehat B = \widehat C \Rightarrow \dfrac{1}{2}\widehat B = \dfrac{1}{2}\widehat C \Rightarrow \widehat {ICB} = \widehat {IBC}\)

Therefore, triangle IBC is isosceles at I (the two base angles are congruent).

\( \Rightarrow IB = IC\)(corresponding edge )

Because EC = BF (sentence b) and IB = IC

\( \Rightarrow \) EC – IC = BF – BI

\( \Rightarrow \) EI = FI

\( \Rightarrow \Delta IEF\) weighed at I

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### Solving problems 5 pages 63 Math textbook 7 Creative horizon volume 2 – CTST

The body of a clothes hanger has the shape of an isosceles triangle (Figure 17a) as shown in Figure 17b. Let AB = 20 cm; BC = 28 cm and \(\widehat B\)= 35°. Find the measures of the remaining angles and perimeter of triangle ABC.

**Detailed instructions for solving Lesson 5**

**Solution method**

Apply isosceles triangle property to find remaining angles and sides

**Detailed explanation**

Since triangle ABC is isosceles at A

\( \Rightarrow \) AB = AC ( isosceles triangle theorem ) = 20 cm

\( \Rightarrow \) Perimeter of triangle ABC = AB + AC + BC = 20 + 20 + 28 = 68 cm

Since ABC is an isosceles triangle at A \( \Rightarrow \widehat B = \widehat C\) ( 2 base angles ) = 35°

According to the theorem, sum of 3 angles in a triangle = 180°

\( \Rightarrow \widehat A = {180^o} – \widehat B – \widehat C = {180^o} – {35^o} – {35^o} = {110^o}\)

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### Solve problem 6 page 63 Math textbook 7 Creative horizon volume 2 – CTST

A triangular window frame of the design shown in Figure 18a is redrawn as shown in Figure 18b.

a) Indicates \(\widehat {{A_1}}\)\( = {42^o}\). Calculate the measure of \(\widehat {{M_1}}\),\(\widehat {{B_1}}\),\(\widehat {{M_2}}\).

b) Prove MN // BC, MP // AC.

c) Prove that the four isosceles triangles AMN, MBP, PMN, and NPC are congruent.

**Detailed instructions for solving Lesson 6**

**Solution method**

Using properties of isosceles triangles

**Detailed explanation**

a) We see that triangle AMN is isosceles at A because AM = AN

\( \Rightarrow \widehat {{M_1}} = ({180^o} – \widehat {{A_1}}):2 = ({180^o} – {42^o}):2 = {69^o }\)

We see triangle PMN = triangle AMN ( ccc )

\( \Rightarrow \widehat {{M_1}} = \widehat {PMN} = {69^o}\) (corresponding angle )

Which \( \Rightarrow \widehat {{M_1}} + \widehat {{M_2}} + \widehat {PMN} = {180^o}\)( complementary angles )

\( \Rightarrow \widehat {{M_2}} = {180^o} – {69^o} – {69^o} = {42^o}\)

And triangle MPB is isosceles at M because MB = MP, so

\( \Rightarrow \widehat {{B_1}} = \widehat {MPB}\)

Apply the sum of the three angles theorem in a triangle

\( \Rightarrow \widehat {{B_1}} = ({180^o} – {42^o}):2 = {69^o}\)

b) We see that \(\widehat {{B_1}}\) and \(\widehat {{M_1}}\) are isotopic and equal, so

\( \Rightarrow \)MN⫽BC

Since triangle PMN = triangle AMN, we have

\( \Rightarrow \widehat {{M_1}} = \widehat {ANM} = \widehat {PMN} = \widehat {MNP}\)( due to 2 isosceles and congruent triangles )

Where \(\widehat {MNA}\) and \(\widehat {PMN}\) are staggered in

\( \Rightarrow \)MP⫽AC

c) We have \(\Delta AMN = \Delta PMN = \Delta MBP(c – g – c)\)(1)

Because MP⫽AC (proven above)

\( \Rightarrow \widehat {MPN} = \widehat {PNC}\) ( 2 staggered angles in ) =\({42^o}\)

\( \Rightarrow \Delta MPN = \Delta NCP(c – g – c)\)(2)

From (1) and (2) \( \Rightarrow \) 4 isosceles triangles AMN, MBP, PMN, NCP are equal

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