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**SBT Solution Lesson 3: Isosceles triangle (C8 SBT Math 7 Horizon)**

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### Solve problem 1 page 49 SBT Math 7 Creative horizon episode 2 – CTST

Given triangle MNP is isosceles at M. Name the sides, base, vertex angle, and base angle of that isosceles triangle.

**Detailed instructions for solving Lesson 1**

**Solution method**

Use the definition of an isosceles triangle to name the sides, base, top angle, and base angle.

**Detailed explanation**

Triangle MNP bound at M has: side sides MN and MP, base side NP, top angle \(\widehat M\), base angle \(\widehat N\) and \(\widehat P \)

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### Solve problem 2 page 49 SBT Math 7 Creative horizon episode 2 – CTST

a) Is a triangle with two angles equal to \({60^o}\) a triangle or not? Find the remaining angle of this triangle.

b) Is a triangle with two angles equal to \({45^o}\) an isosceles triangle or not? Find the remaining angles of this triangle.

**Detailed instructions for solving Lesson 2**

**Solution method**

Use the definition of an isosceles triangle.

– Use the theorem that the sum of the three angles in a triangle is equal to \({180^o}\) to calculate the remaining angles of the triangle.

**Detailed explanation**

a) The remaining angle is equal to: \({180^o} – {2.60^o} = {60^o}\). This triangle is both an equilateral triangle and an isosceles triangle at its three vertices.

b) The remaining angle is equal to: \({180^o} – {2.45^o} = {90^o}\). This triangle is both an isosceles triangle and a right triangle and is called an isosceles right triangle for short.

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### Solution 3 page 49 SBT Math 7 Creative horizon episode 2 – CTST

In Figure 6, calculate angle B and angle C knowing \(\widehat {{A^{}}} = {138^o}\)

**Detailed instructions for solving Lesson 3**

**Solution method**

Use the isosceles triangle property to find the measure of the angle you want to find

**Detailed explanation**

We have: \(\widehat B = \widehat C = \frac{{{{180}^o} – {{80}^o}}}{2} = {50^o}\)(because of triangle ABC weigh at A)

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### Solution 4 page 49 SBT Math 7 Creative horizon episode 2 – CTST

Given figure 7, know AB = AC and BE is the bisector of \(\widehat {ABC}\), CF is the bisector of angle \(\widehat {ACB}\). Prove that:

a) \(\Delta ABE = \Delta ACF\)

b) Isosceles triangle OEF

**Detailed instructions for solving Lesson 4**

**Solution method**

– Use the property of bisectors to show that two angles are congruent from proving two triangles are congruent.

– Proving OE = OF should deduce triangle OEF isosceles

**Detailed explanation**

a) we have AB = AC, so triangle ABC is isosceles at A, so \(\widehat B = \widehat C\)

Otherwise: \(\widehat {FCA} = \frac{{\widehat C}}{2}\) (because CF is the bisector of angle \(\widehat {ACB}\))

\(\widehat {EBA} = \frac{{\widehat B}}{2}\) (because BE is the bisector of \(\widehat {ABC}\))

So \(\widehat {FCA} = \widehat {EBA}\)

Consider triangle ACF and triangle ABE with:

\(\widehat {{A^{}}}\)general

AC = AB

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\(\widehat {FCA} = \widehat {EBA}\)

Reason: \(\Delta ACF = \Delta ABE(g – c – g)\)

b) We have: \(\Delta ACF = \Delta ABE(g – c – g)\) infer: BE = CF (1)

We have an isosceles triangle OBC at O, so OB = OC (2)

From (1) and (2) deduce BE – OB = CF – OC, so OE = OF

So triangle OEF is isosceles at O.

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### Solution 5 page 49 SBT Math 7 Creative horizon episode 2 – CTST

Given triangle MEF isosceles at M, there are \(\widehat M = {80^o}\)

a) Calculate \(\widehat E{,^{}}\widehat F\)

b) Let N, P be the midpoints of ME, MF, respectively. Prove that triangle MNP is isosceles.

c) Prove that NP // EF

**Detailed instructions for solving Lesson 5**

**Solution method**

Use the isosceles triangle property to find the measures of the angles

– Prove that MN = MP deduce the triangle MNP is isosceles at M

– Prove two angles \(\widehat {MNP} = \widehat {{\rm{NEF}}}\) infer NP // EF

**Detailed explanation**

a) Since the triangle MEF is isosceles at M, \(\widehat E = \widehat F = \frac{{{{180}^o} – {{80}^o}}{2} = {50^o} \)

b) We have an isosceles triangle MEF at M so ME = MF.

Derive: \(MN = \frac{{ME}}{2} = \frac{{MF}}{2} = MP\)

So triangle MNP is isosceles at M.

c) In isosceles triangle MNP we have: \(\widehat N = \widehat P = \frac{{{{180}^o} – {{80}^o}}}{2} = {50^o} \)

so \(\widehat {MNP} = \widehat {{\rm{NEF}}} = {50^o}\)

Derive NP // EF (since two isotopic angles are equal)

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### Solution 6 page 50 SBT Math 7 Creative horizon episode 2 – CTST

Let ABC be an isosceles right triangle at A. The bisector of angle B intersects AC at N, the bisector of angle C intersects AB at M. Let O be the intersection of BN and CM.

a) Calculate the measure of angles OBC, OCB.

b) Prove that triangle OBC is isosceles.

c) Calculate the measure of angle BOC.

**Detailed instructions for solving Lesson 6**

**Solution method**

Use the bisector of an angle to calculate the angle measure.

– Prove that \(\widehat {OBC} = \widehat {OCB}\) deduces triangle OBC isosceles at O.

– Use the theorem that the sum of the three angles in a triangle is equal to \({180^o}\) to calculate the remaining angles of the triangle.

**Detailed explanation**

a) Since triangle ABC is right-angled at A, \(\widehat B = \widehat C = {45^o}\)

We have: \(\widehat {OBC} = \widehat {OCB} = \frac{{{{45}^o}}}{2} = 22,{5^o}\)

b) Triangle OBC has \(\widehat {OBC} = \widehat {OCB}\) so triangle OBC is isosceles at O.

c) we have: \(\widehat {BOC} = {180^o} – \left( {\widehat {OBC} + \widehat {OCB}} \right) = {180^o} – {45^o} = {135^o}\)

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