# 2.5: Isosceles Triangles

- Page ID
- 34128

In Section 1.6, we defined a triangle to be isosceles if two of its sides are equal. Figure \(\PageIndex{1}\) shows an isosceles triangle \(\triangle ABC\) with \(AC=BC\). In \(\triangle ABC\) we say that \(\angle A\) is opposite side \(BC\) and \(\angle B\) is opposite side \(AC\).

The most important fact about isosceles triangles is the following:

If two sides of a triangle are equal the angles opposite these sides are equal.

Theorem \(\PageIndex{1}\) means that if \(AC = BC\) in \(\triangle ABC\) then \(\angle A = \angle B\).

Find \(x\):

**Solution**

\(AC = BC\) so \(\angle A = \angle B\). Therefore, \(x = 40\).

Answer: \(x = 40\).

In \(\triangle ABC\) if \(AC = BC\) then side \(AB\) is called the **base **of the triangle and \(\angle A\) and \(\angle B\) are called the base angles. Therefore Theorem \(\PageIndex{1}\) is sometimes stated in the following way: "The base angles of an isosceles triangle are equal,"

Proof of Theorem \(\PageIndex{1}\): Draw \(CD\), the angle bisector of \(\angle ACB\) (Figure \(\PageIndex{2}\)). The rest of the proof will be presented in double-column form. We have given that \(AC = BC\) and \(\angle ACD = \angle BCD\). We must prove \(\angle A = \angle B\).

Statements | Reasons |
---|---|

1. \(AC = BC\). | 1. Given, \(\triangle ABC\) is isosceles. |

2. \(\angle ACD = \angle BCD\). | 2. Given, \(CD\) is the angle bisector of \(\angle ACB\). |

3. \(CD = CD\). | 3. Identity. |

4. \(\triangle ACD \cong \triangle BCD\). | 4. \(SAS = SAS\): \(AC, \angle C, CD\) of \(\triangle ACD = BC\), \(\angle C, CD\) of \(\triangle BCD\). |

5. \(\angle A = \angle B\). | 5. Corresponding angles of congruent trianglesare equal. |

Find \(x, \angle A, \angle B\) and \(\angle C\):

**Solution**

\(\angle B = \angle A = 4x + 5^{\circ}\) by Theorem \(\PageIndex{1}\). We have

\[\begin{array} {rcl} {\angle A + \angle B + \angle C} & = & {180^{\circ}} \\ {4x + 5 + 4x + 5 + 2x - 10} & = & {180} \\ {10x} & = & {180} \\ {x} & = & {18} \end{array}\]

\(\angle A = \angle B = 4x + 5^{\circ} = 4(18) + 5^{\circ} = 72 + 5^{\circ} = 77^{\circ}\).

\(\angle C = 2x - 10^{\circ} = 2(18) - 10^{\circ} = 36 - 10^{\circ} = 26^{\circ}\).

**Check**

**Answer**

\(x = 18\), \(\angle A = 77^{\circ}\), \(\angle B = 77^{\circ}\), \(\angle C = 26^{\circ}\).

In Theorem \(\PageIndex{1}\) we assumed \(AC = BC\) and proved \(\angle A = \angle B\). We will now assume \(\angle A = \angle B\) and prove \(AC = BC\). '1ihen the assumption and conclusion of a statement are interchanged the result is called the converse of the original statement.

If two angles of a triangle are equal the sides opposite these angles are equal.

If Figure 4, if \(\angle A = \angle B\) then \(AC = BC\).

Find \(x\)

**Solution**

\(\angle A = \angle B\) so \(x = AC = BC = 9\) by Theorem \(\PageIndex{2}\).

**Answer**

\(x = 9\).

Proof of Theorem \(\PageIndex{2}\): Draw \(CD\) the angle bisector of \(\angle ACB\) (Figure \(\PageIndex{5}\)). We have \(\angle ACD = \angle BCD\) and \(\angle A = \angle B\). We must prove \(AC = BC\).

Statements | Reasons |
---|---|

1. \(\angle A = \angle B\). | 1. Given. |

2. \(\angle ACD = \angle BCD\). | 2. Given. |

3. \(CD = CD\). | 3. Identity. |

4. \(\triangle ACD \cong \triangle BCD\). | 4. \(AAS = AAS\): \(\angle A, \angle C, CD\) of \(\triangle ACD = \angle B\), \(\angle C\), \(CD\) of \(triangle BCD\). |

5. \(AC = BC\). | 5. Corresponding sides of congruent triangles are equal |

The following two theorems are **corollaries** (immediate consequences) of the two preceding theorems:

An equilateral triangle is equiangular.

In Figure \(\PageIndex{7}\), if \(AB = AC = BC\) then \(\angle A = \angle B = \angle C\).

**Proof**-
\(AC = BC\) so by Theorem \(\PageIndex{1}\) \(\angle B = \angle C\). Therefore \(\angle A = \angle B = \angle C\).

Since the sum of the angle is \(180^{\circ}\) we must have in fact that \(\angle A = \angle B = \angle C = 60^{\circ}\).

An equiangular triangle is equilateral.

In Figure \(\PageIndex{8}\), if \(\angle A = \angle B = \angle C\) then \(AB = AC = BC\).

**Proof**-
\(\angle A = \angle B\) so by Theorem \(\PageIndex{2}\), \(AC = BC\), \(\angle B = \angle C\) by Theorem \(\PageIndex{2}\), \(AB = AC\). Therefore \(AB = AC = BC\).

Find \(x\), \(y\) and \(AC\):

**Solution**

\(\triangle ABC\) is equiangular and so by Theorem \(\PageIndex{4}\) is equilateral.

Therefore \(\begin{array} {rcl} {AC} & = & {AB} \\ {x + 3y} & = & {7x - y} \\ {x - 7x + 3y + y} & = & {0} \\ {-6x + 4y} & = & {0} \end{array}\) and \(\begin{array} {rcl} {AB} & = & {BC} \\ {7x - y} & = & {3x + 5} \\ {7x - 3xy - y} & = & {5} \\ {4x - y} & = & {5} \end{array}\)

We have a system of two equations in two unknowns to solve:

Check:

Answer: \(x = 2\), \(y = 3\), \(AC = 11\).

Theorem \(\PageIndex{1}\), the isosceles triangle theorem, is believed to have first been proven by Thales (c. 600 B,C,) - it is Proposition 5 in Euclid's **Elements**. Euclid's proof is more complicated than ours because he did not want to assume the existence of an angle bisector, Euclid's proof goes as follows:

Given \(\triangle ABC\) with \(AC = BC\) (as in Figure \(\PageIndex{1}\) at the beginning of this section), extend \(CA\) to \(D\) and \(CB\) to \(E\) so that \(AD = BE\) (Figure \(\PageIndex{9}\)). Then \(\triangle DCB \cong \triangle ECA\) by \(SAS = SAS\). The corresponding sides and angles of the congruent triangles are equal, so \(DB = EA\), \(\angle 3 = \angle 4\) and \(\angle 1 + \angle 5 = \angle 2 + \angle 6\). Now \(\triangle ADB \cong \triangle BEA\) by \(SAS = SAS\). This gives \(\angle 5 = \angle 6\) and finally \(\angle 1 = \angle 2\).

This complicated proof discouraged many students from further study in geometry during the long period when the **Elements** was the standard text, Figure \(\PageIndex{9}\) resembles a bridge which in the Middle Ages became known as the "bridge of fools," This was supposedly because a fool could not hope to cross this bridge and would abandon geometry at this point.

## Problems

For each of the following state the theorem(s) used in obtaining your answer.

1. Find \(x\):

2. Find \(x\), \(\angle A\), and \(\angle B\):

3. Find \(x\):

4. Find \(x\), \(AC\), and \(BC\):

5. Find \(x\):

6. Find \(x\):

7. Find \(x, \angle A, \angle B\), and \(\angle C\):

8. Find \(x, \angle A, \angle B\), and \(\angle C\):

9. Find \(x, AB, AC\), and \(BC\):

10. Find \(x, AB, AC\), and \(BC\):

11. Find \(x, y\), and \(AC\):

12. Find \(x, y\), and \(AC\):

13. Find \(x\):

14. Find \(x, y\), and \(z\):