Properties of Isosceles Triangles

Anisosceles triangleis a triangle that has (at least) two equal side lengths. If all three side lengths are equal, the triangle is alsoequilateral. Isosceles triangles are very helpful in determining unknown angles.

In an isosceles triangle, the two equal sides are calledlegs, and the remaining side is called thebase. The angle opposite the base is called thevertex angle, and the point associated with that angle is called theapex. The two equal angles are called theisosceles angles.

If the triangle is alsoequilateral, any of the three sides can be considered the base.

Because angles opposite equal sides are themselves equal, an isosceles triangle has two equal angles (the ones opposite the two equal sides). Thus, given two equal sides and a single angle, the entire structure of the triangle can be determined. Likewise, given two equal angles and the length of any side, the structure of the triangle can be determined.

Determining the area can be done with only a few pieces of information (namely, 3):

• A triangle with base $b$and equal side $\ell$has area $\frac{b\sqrt{4\ell^2-b^2}}{4}$.
• A triangle with base $b$and isosceles angle $\theta$has area $\frac{b^2\tan \theta}{4}$, as the length of the altitude to the base is $\压裂{b \ tan \θ}{2}$. Alternatively, if the apex angle has measure $\alpha$, the area is $\frac{b^2}{4}\cot\frac{\alpha}{2}$.
• A triangle with equal side $\ell$and isosceles angle $\theta$has area $\frac{\ell^2\sin 2\theta}{2}$, as the vertex angle has measure $180^{\circ}-2\theta$, and $\sin(180^{\circ}-2\theta)=\sin 2\theta$. Alternatively, if the vertex angle has measure $\alpha$, the area is $\压裂{α罪\魔法^ 2 \ \}{2}$.

The altitude to the base also satisfies important properties:

• The altitude to the base is the perpendicular bisector of the base.
• The altitude to the base is the angle bisector of the vertex angle.
• The altitude to the base is theline of symmetryof the triangle.
• The altitude to the base is the median from the apex to the base.

This means that theincenter,circumcenter,centroid, andorthocenterall lie on the altitude to the base, making the altitude to the base theEuler lineof the triangle.

It is immediate that any $n$-sidedregular polygoncan be decomposed into $n$isosceles triangles, where each triangle contains two vertices and the center of the polygon.

$\begin{aligned} R &= \frac{S}{2 \sin{\frac{\phi}{2}}} \\ S &= 2 R \sin{\frac{\phi}{2}} \\ r &= R \cos{\frac{\phi}{2}} \\ Area &= \frac{1}{2} R^2 \sin{\phi} \end{aligned}$

A regular $n$-gon is composed of $n$isosceles congruent triangles. Any isosceles triangle is composed of two congruent right triangles as shown in the sketch. These right triangles are very useful in solving $n$-gon problems. The relation given could be handy. Apart from the above-mentioned isosceles triangles, there could be many other isoceles triangles in an $n$-gon.

$n \times \phi =2 \pi = 360^{\circ}.$

More interestingly, any triangle can be decomposed into $n$isosceles triangles, for any positive integer $n \geq 4$. The picture to the right shows a decomposition of a 13-14-15 triangle into four isosceles triangles.

Isosceles triangle $ABC$has $AB=AC$and $\angle BAC=40^{\circ}$. What is the measure of $\angle ABC$?

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Because $AB=AC$, we know that $\angle ABC=\angle ACB$. Additionally, the sum of the three angles in a triangle is $180^{\circ}$, so $\angle ABC+\angle ACB+\angle BAC=2\angle ABC+\angle BAC=180^{\circ}$, and since $\angle BAC=40^{\circ}$, we have $2\angle ABC=140^{\circ}$. Thus $\angle ABC=70^{\circ}$. $_\square$

在等腰三角形 $ABC$, the measure of $\angle ABC$is $50^{\circ}$. What are the possible measures of $\angle BAC$?

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There are three possible cases:

• Case 1: $\angle ABC=\angle BAC$. This is easy to deal with, as then $\angle BAC=50^{\circ}$immediately.
• Case 2: $\angle ABC=\angle BCA$. Since the angles of a triangle add up to $180^{\circ}$, we know that $50^{\circ}+50^{\circ}+\angle BAC=180^{\circ}$, and so $\angle BAC=80^{\circ}$.
• Case 3: $\angle BAC=\angle BCA$. Again, the angles of a triangle add up to $180^{\circ}$, so $50^{\circ}+2\angle BAC=180^{\circ}$, and so $\angle BAC=65^{\circ}$.

Therefore, the possible values of $\angle BAC$are $50^{\circ}, 65^{\circ}$, and $80^{\circ}$. $_\square$

In the above figure, $AD=DC=CB$and the measure of $\angle DAC$is $40^{\circ}$. What is the measure of $\angle DCB$?

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From the given condition, both $\三角形ADC$and $\triangle DCB$are isosceles triangles.

In $\三角形ADC$, $\angle DCA=\angle DAC=40^{\circ}$, implying $\angle CDB=40^{\circ}+40^{\circ}=80^{\circ}$by theexterior angle of a triangle.

In $\triangle DCB$, $\angle CBD=\angle CDB=80^{\circ}$, implying $\angle DCB=180^{\circ}-80^{\circ}-80^{\circ}=20^{\circ}$by theangle sum of a triangle. $_\square$

• Properties of equilateral triangles