The isosceles triangle

An isosceles triangle is a triangle with two equal sides.

• The legs are the two equal sides: here $\displaystyle AB$ and $\displaystyle AC$.
• The base is the third side: $\displaystyle BC$.
• The apex angle is the angle between the legs: $\displaystyle \angle A$.
• The base angles are the two angles next to the base: $\displaystyle \angle B$ and $\displaystyle \angle C$.

Theorem: in an isosceles triangle, the base angles are equal.

• In triangle $\displaystyle ABC$, if $\displaystyle {AB}={AC}$, then $\displaystyle {\angle B}={\angle C}$.
• Each base angle faces a leg: $\displaystyle \angle C$ faces $\displaystyle AB$, and $\displaystyle \angle B$ faces $\displaystyle AC$.

Given: $\displaystyle {AB}={AC}$.
Prove: $\displaystyle {\angle B}={\angle C}$.

The proof below draws $\displaystyle AD$, the bisector of angle $\displaystyle A$, and shows that triangles $\displaystyle ABD$ and $\displaystyle ACD$ are congruent by SAS:

In an isosceles triangle, the bisector of the apex angle is also the median to the base and the altitude to the base.

The proof found $\displaystyle \triangle ABD\cong\triangle ACD$. Two more facts follow from the congruence:

• $\displaystyle {BD}={CD}$ – corresponding sides, so $\displaystyle AD$ is the median to the base;
• $\displaystyle {\angle ADB}={\angle ADC}$ – corresponding angles. They are adjacent angles, together $\displaystyle 180^{\circ}$, so each is $\displaystyle 90^{\circ}$: $\displaystyle AD$ is the altitude to the base.

From the vertex $\displaystyle A$ there is one median to the base and one altitude to the base, so all three – the angle bisector, the median and the altitude – are the same segment.

The base angles are equal, so one angle of an isosceles triangle is enough to find the other two.

• The apex angle is $\displaystyle 40^{\circ}$: the two base angles together are $\displaystyle {180^{\circ}-40^{\circ}}={140^{\circ}}$, and each is $\displaystyle {140^{\circ}:2}={70^{\circ}}$.
• A base angle is $\displaystyle 65^{\circ}$: the other is $\displaystyle 65^{\circ}$ too, and the apex angle is $\displaystyle {180^{\circ}-2\cdot65^{\circ}}={50^{\circ}}$.

In triangle $\displaystyle ABC$, $\displaystyle {AB}={AC}$.
Find the angles, and give the reasons.

$\displaystyle {\angle A}={100^{\circ}}$. Find the angle $\displaystyle B$ and the angle $\displaystyle C$.

$\displaystyle {\angle B}={72^{\circ}}$. Find the angle $\displaystyle A$.

$\displaystyle AD$ is the median to the base $\displaystyle BC$, and $\displaystyle {\angle BAD}={25^{\circ}}$.
Find the angle $\displaystyle BAC$, the angle $\displaystyle B$ and the angle $\displaystyle ADB$.

In triangle $\displaystyle ABC$, $\displaystyle {AB}={AC}$.
The angle $\displaystyle A$ is $\displaystyle x$ degrees, and the angle $\displaystyle B$ is $\displaystyle 2x+10$ degrees.
Write an equation, and find $\displaystyle x$.

An equilateral triangle is a triangle with three equal sides. Each of its angles is $\displaystyle 60^{\circ}$:

• $\displaystyle {AB}={AC}$, so $\displaystyle {\angle B}={\angle C}$ – the base angles when the base is $\displaystyle BC$;
• $\displaystyle {BA}={BC}$, so $\displaystyle {\angle A}={\angle C}$ – the base angles when the base is $\displaystyle AC$;
• the three angles are equal, so each is $\displaystyle {180^{\circ}:3}={60^{\circ}}$.

In triangle $\displaystyle ABC$, $\displaystyle {AB}={AC}$.
The points $\displaystyle D$ and $\displaystyle E$ are on the base $\displaystyle BC$, so that $\displaystyle {BD}={CE}$.
Prove: $\displaystyle {AD}={AE}$.

In triangle $\displaystyle ABC$, $\displaystyle {AB}={AC}$.
$\displaystyle BD$ is the altitude to the leg $\displaystyle AC$, and $\displaystyle CE$ is the altitude to the leg $\displaystyle AB$.
Prove: $\displaystyle {BD}={CE}$.