Math 7 Chapter 2 Lesson 6: Isosceles triangle
1. Summary of theory
1.1. Define
An isosceles triangle is a triangle with two equal sides.
Eg: Consider triangle ABC with AB = AC, so triangle ABC is isosceles at A.
1.2. Nature
In an isosceles triangle the two base angles are congruent.
 If a triangle has two equal angles, then the triangle is isosceles
 An isosceles right triangle is a right triangle with two equal right angles.
Eg: Consider triangle ABC with \(\widehat B = \widehat C\) so triangle ABC is isosceles at A.
1.3. equilateral triangle
Define: An equilateral triangle is a triangle with three equal sides.
Eg: An equilateral triangle ABC has AB = AC = BC.
Consequences:
 In an equilateral triangle, each angle is equal to \({60^0}\)
 If a triangle has three equal angles, then the triangle is an equilateral triangle.
 If an isosceles triangle has an angle equal to \({60^0}\) then the triangle is an equilateral triangle.
Eg:
 An equilateral triangle ABC has \(\widehat A = \widehat B = \widehat C = 60^\circ \)
 If triangle ABC has \(\widehat A = \widehat B = \widehat C\) then triangle ABC is an equilateral triangle.
 If triangle ABC is isosceles \(\widehat A = 60^\circ \) then triangle ABC is equilateral.
2. Illustrated exercise
Question 1: Let ABC be an isosceles triangle at A with \(\widehat A = {50^0}\)
a. Calculate \(\widehat B,\,\,\widehat C\)
b. Take point D on side AB, point E on side AC such that AD = AE. Prove that DE // BC.
Solution guide
a. We have:
\(\begin{array}{l}\widehat B = \,\,\widehat C = \frac{{{{180}^0} – \widehat A}}{2} = \frac{{{{180 }^0} – {{50}^0}}}{2}\\ = \widehat B = \,\,\widehat C = {65^0}\,{\,^{(1)}}\ end{array}\)
b. AD = AE so \(\Delta ADE\) bounds at A
Derive \(\,\widehat {ADE} = \frac{{{{180}^0} – \widehat A}}{2} = \frac{{{{180}^0} – {{50}^ 0}}}{2} = {65^0}\,{\,^{(2)}}\)
From (1) and (2) deduce \(\widehat B = \widehat {ADE}\)
So DE // BC (two isotopic angles are equal)
Verse 2: Let an isosceles triangle at A. Let D be the mid point of AC, let E be the mid point of AB. Compare the lengths BD and CE.
Solution guide
See drawing:
Method 1: \(\Delta ABD\) and \(\Delta ACE\) have:
AB = AC (gt)
\(\widehat A\) common
So \(\Delta ABD = \Delta ACE\,\,(cgc)\)
Derive BD = CE.
Method 2: \(\Delta BDC\) and \(\Delta CEB\) have
CD = BE (gt)
\(\widehat B = \widehat {C\,}\,(gt)\)
BC common edge
So \(\Delta BDC = \Delta CEB\,\,\,(cgc)\)
Derive BD = CE
Question 3: Let \(\Delta ABC\) is isosceles at A and there is \(\widehat B = 2\widehat A\) bisector of angle B that intersects AC at D.
a. Calculate the angles of \(\Delta ABC\)
b. Prove DA = DB
c. Prove DA = BC
Solution guide
a. We have \(\widehat {A\,} + \widehat {B\,} + \widehat {C\,} = {180^0}\)
where \(\Delta ABC\)weighs at A, there \(\widehat B = 2\widehat A\), so:
\(\widehat {A\,} + 2\widehat {A\,} + \widehat {A\,} = {180^0}\)
Replace \(5\widehat {A\,} = {180^0} \Rightarrow \widehat {A\,} = {36^0}\)
So \(\widehat {B\,} = \widehat {C\,} = 2\widehat {A\,} = {72^0}\)
b. We have: \(\widehat {DBA} = \frac{1}{2}\widehat B = {36^0}\) (BD bisector \(\widehat B\))
where \(\widehat {A\,} = {36^0}\) so \(\widehat {A\,} = \widehat {DBA}\)
It follows that \(\Delta ABD\) is balanced at D
So \(DA = DB{\,^{\,(1)}}\)
c. We have: \(\widehat {BDC}\) is the exterior angle at D of \(\Delta ABD\) so
\(\widehat {BDC} = \widehat {DBA} + \widehat A = {36^0} + {36^0} = {72^0}\)
Which \(\widehat C = {72^0}\) deduces \(\Delta DBC\) equals at B
So BD = BC (2)
From (1) and (2) deduce AD = BC.
3. Practice
3.1. Essay exercises
Question 1: Given two parallel lines x’x and y’y and a line that intersects x’x at M and y’y at N. On the line y’y take two points E, F on either side of N such that NE=NF=NM. Prove:
a. ME, MF are bisectors of two angles \(\widehat {xMN}\) and \(\widehat {x’MN}\)
b. \(\Delta M{\rm{EF}}\) is a right triangle
Verse 2: Let ABC be an isosceles triangle (AB=AC) on the opposite ray of BC take point D and on the opposite ray of CB take point E such that CE = BD. If A with D and A with E.
a. Compare \(\widehat {ABD}\) and \(\widehat {ACE}\)
b. Prove that \(\Delta ADE\) is balanced.
Question 3: Given \(\Delta ABD,\,\widehat B = 2\widehat D\), line \(AH \bot BD\,\;(H \in BD)\)
On the opposite ray of ray BA take BE = BH. The line EH intersects ED at F. Prove: FH = FA = FD.
3.2. Multiple choice exercises
Question 1: Choose the wrong sentence:
A. An equilateral triangle has three equal angles and is equal to 60^{0}
B. An equilateral triangle has three equal sides
C. Isosceles triangle is an equilateral triangle
D. An equilateral triangle is an isosceles triangle
Verse 2: The two acute angles of a right triangle are congruent and equal to:
A. 30^{0}
B. 45^{0}
C. 60^{0}
D. 90^{0}
Question 3: Let ABC be an isosceles triangle at A. Which of the following statements is false:
A. \(\widehat B = \widehat C\)
B. \(\widehat C = \frac{{{{180}^0} – \widehat A}}{2}\)
C. \(\widehat A = {180^0} – 2\widehat C\)
D. \(\widehat B \ne \widehat C\)
Question 4: An isosceles triangle whose vertex angle is 64^{0} then the measure of the angle at the base is:
A. 54^{0}
B. 58^{0}
C. 72^{0}
D. 90^{0}
Question 5: An isosceles triangle whose base angle is 70^{0} then the measure of the angle at the vertex is:
A. 54^{0}
B. 63^{0}
C. 70^{0}
D. 40^{0}
4. Conclusion
Through this lesson, you should achieve the following goals:

Definition and properties of an isosceles triangle.

Definition and properties of an equilateral triangle.
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