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# 2.5: Isosceles Triangles

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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 $$AC$$. Figure $$\PageIndex{1}$$: $$\triangle ABC$$ is isosceles with AC = BC.

The most important fact about isosceles triangles is the following:

Theorem $$\PageIndex{1}$$

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 $$\angle A = \angle B$$.

Example $$\PageIndex{1}$$

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 $$\angle A = \angle B$$. Figure $$\PageIndex{2}$$: Draw $$CD$$, the angle bisector of $$\angle ACB$$.
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.

Example $$\PageIndex{2}$$

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.

Theorem $$\PageIndex{2}$$: The Converse of Theorem $$\PageIndex{1}$$)

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

If Figure 4, if $$\angle A = \angle B$$ then $$AC = BC$$. Figure $$\PageIndex{4}$$. $$\angle A = \angle B$$

Example $$\PageIndex{3}$$

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 $$AC = BC$$. Figure $$\PageIndex{5}$$. Draw $$CD$$, the angle bisector of $$\angle ACB$$.
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:

Theorem $$\PageIndex{3}$$

An equilateral triangle is equiangular.

In Figure $$\PageIndex{7}$$, if $$AB = AC = BC$$ then $$\angle A = \angle B = \angle C$$. Figure $$\PageIndex{7}$$: $$\angle ABC$$ is equilateral.
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 $$\angle A = \angle B = \angle C = 60^{\circ}$$.

Theorem $$\PageIndex{4}$$: The Converse of Theorem $$\PageIndex{3}$$)

An equiangular triangle is equilateral.

In Figure $$\PageIndex{8}$$, if $$\angle A = \angle B = \angle C$$ then $$AB = AC = BC$$. Figure $$\PageIndex{8}$$. $$\angle ABC$$ is equiangular.
Proof

$$\angle A = \angle B$$ so by Theorem $$\PageIndex{2}$$, $$AC = BC$$, $$\angle B = \angle C$$ by Theorem $$\PageIndex{2}$$, $$AB = AC$$. Therefore $$AB = AC = BC$$.

Example $$\PageIndex{4}$$

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: $$x = 2$$, $$y = 3$$, $$AC = 11$$.

Historical Note

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 $$\angle 1 = \angle 2$$. Figure $$\PageIndex{9}$$: The "bridge of fools".

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 $$z$$: • Was this article helpful?