The isosceles triangle: bisector, median and altitude

In eighth grade you learned: in an isosceles triangle, the bisector of the apex angle, the median to the base and the altitude to the base coincide – they are the same segment.

Today the other way round: it is enough that two of the three coincide, and the triangle is isosceles.

In triangle $\displaystyle ABC$, the segment $\displaystyle AD$ goes from the vertex $\displaystyle A$ to the side facing it, $\displaystyle BC$. It can be:

• an angle bisector – when it divides angle $\displaystyle A$ into two equal angles: $\displaystyle {\angle BAD}={\angle CAD}$;
• a median – when the point $\displaystyle D$ is the midpoint of the side: $\displaystyle {BD}={CD}$;
• an altitude – when it is perpendicular to the side: $\displaystyle AD\perp BC$.

Theorem: if two of the angle bisector, the median and the altitude from the same vertex of a triangle coincide, the triangle is isosceles.

These are three theorems, one for each pair. In triangle $\displaystyle ABC$ the point $\displaystyle D$ is on the side $\displaystyle BC$, and $\displaystyle {AB}={AC}$ if the segment $\displaystyle AD$ is:

• an angle bisector and an altitude;
• an angle bisector and a median;
• a median and an altitude.

Given: $\displaystyle AD\perp BC$ and $\displaystyle {\angle BAD}={\angle CAD}$.
Prove: $\displaystyle {AB}={AC}$.

The proof below shows that triangles $\displaystyle ABD$ and $\displaystyle ACD$ are congruent by ASA: the two angles at $\displaystyle AD$ are equal, and $\displaystyle AD$ is a common side.

Given: $\displaystyle {\angle BAD}={\angle CAD}$ and $\displaystyle {BD}={CD}$.
Prove: $\displaystyle {AB}={AC}$.

The proof below needs a construction: it extends $\displaystyle AD$ past $\displaystyle D$ to the point $\displaystyle E$, so that $\displaystyle {DE}={AD}$, and joins $\displaystyle E$ to $\displaystyle C$:

• Triangle $\displaystyle ECD$ is congruent to triangle $\displaystyle ABD$, so $\displaystyle {EC}={AB}$;
• triangle $\displaystyle ACE$ has two equal angles, so $\displaystyle {EC}={AC}$.

In triangle $\displaystyle ABC$, the altitude $\displaystyle AD$ is also the bisector of angle $\displaystyle A$.
$\displaystyle {\angle BAD}={25^{\circ}}$ and $\displaystyle {AB}={9}$.

• $\displaystyle AD$ is an altitude and an angle bisector, so the triangle is isosceles: $\displaystyle {AC}={AB}={9}$;
• angle $\displaystyle B$ of triangle $\displaystyle ABD$ is $\displaystyle {180^{\circ}-90^{\circ}-25^{\circ}}={65^{\circ}}$;
• the base angles are equal, so $\displaystyle {\angle C}={65^{\circ}}$ too, and the apex angle is $\displaystyle {\angle BAC}={2\cdot25^{\circ}}={50^{\circ}}$.

Prove the third theorem: a median that is also an altitude.

Given: $\displaystyle {BD}={CD}$ and $\displaystyle AD\perp BC$.
Prove: $\displaystyle {AB}={AC}$.

Calculate, and give the reason for each step.

In triangle $\displaystyle KLM$, the segment $\displaystyle KN$ is the median to the side $\displaystyle LM$ and also the altitude to it.
$\displaystyle {KL}={11}$ and $\displaystyle {\angle L}={52^{\circ}}$.
Find $\displaystyle KM$ and angle $\displaystyle LKM$.

In triangle $\displaystyle PQR$, the bisector of angle $\displaystyle P$ is also the median to the side $\displaystyle QR$.
$\displaystyle {\angle Q}={35^{\circ}}$.
Find angle $\displaystyle R$ and angle $\displaystyle P$.

In triangle $\displaystyle ABC$, the segment $\displaystyle BD$ bisects angle $\displaystyle B$.
The line through $\displaystyle A$ perpendicular to $\displaystyle BD$ meets $\displaystyle BD$ at the point $\displaystyle F$, and $\displaystyle BC$ at the point $\displaystyle E$.
Prove: $\displaystyle {AB}={BE}$.