The isosceles triangle: the converse theorems

In a converse theorem, what is given and what is to be proved change places.

• The theorem: if $\displaystyle {AB}={AC}$, then $\displaystyle {\angle B}={\angle C}$.
• The converse theorem: if $\displaystyle {\angle B}={\angle C}$, then $\displaystyle {AB}={AC}$.

Careful: the converse of a true theorem is not always true. “If a quadrilateral is a square, all its angles are right” – true; “if all the angles of a quadrilateral are right, it is a square” – false: a long rectangle has four right angles too. So every converse theorem needs a proof of its own.

Theorem: a triangle with two equal angles is isosceles: the sides facing the equal angles are equal.

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

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

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

In triangle $\displaystyle ABC$, $\displaystyle {\angle A}={50^{\circ}}$ and $\displaystyle {\angle B}={65^{\circ}}$.
The sides: $\displaystyle {AB}={2x+3}$ and $\displaystyle {AC}={x+8}$ (in cm).
Find $\displaystyle x$ and the length of the side $\displaystyle AB$. Give the reasons.

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

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

The proof below adds a helping construction: it extends $\displaystyle AD$ beyond $\displaystyle D$ 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 $\displaystyle {EC}={AC}$.

Prove the theorem: an angle bisector that is also an altitude.

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

Prove the 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.
$\displaystyle {KL}={10}$ cm and $\displaystyle {\angle L}={48^{\circ}}$.
Find $\displaystyle KM$ and the 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 the angle $\displaystyle R$ and the angle $\displaystyle P$.