The fourth congruence theorem and the converse of Pythagoras

Theorem: if two sides of one triangle are equal to two sides of another, and the angles opposite the larger of the two sides are equal, the triangles are congruent.

Given: $\displaystyle {AB}={DE}$, $\displaystyle {AC}={DF}$, $\displaystyle {\angle C}={\angle F}$, and $\displaystyle AB>AC$ – the angles $\displaystyle C$ and $\displaystyle F$ are opposite the larger sides.
To prove: $\displaystyle \triangle ABC\cong\triangle DEF$.

The proof below is a proof by contradiction: it supposes that $\displaystyle BC\neq EF$ and shows that this leads to something false.

• Suppose $\displaystyle EF<BC$. Take $\displaystyle G$ on $\displaystyle CB$ with $\displaystyle {CG}={FE}$.
• $\displaystyle \triangle AGC\cong\triangle DEF$ (SAS: $\displaystyle {AC}={DF}$, $\displaystyle {\angle C}={\angle F}$, $\displaystyle {CG}={FE}$), so $\displaystyle {AG}={DE}={AB}$.
• So the triangle $\displaystyle ABG$ is isosceles, and its base angles $\displaystyle \angle ABG$ and $\displaystyle \angle AGB$ are equal – both acute, since two of them fit in one triangle.
• So $\displaystyle {\angle AGC}={180^{\circ}-\angle AGB}$ is obtuse. It is the largest angle of the triangle $\displaystyle AGC$, and the side opposite it is the largest: $\displaystyle {AC>AG}={AB}$.
• That contradicts $\displaystyle AB>AC$. In the same way $\displaystyle EF>BC$ leads to a contradiction, with the roles of the triangles swapped.

So $\displaystyle {BC}={EF}$, and the triangles are congruent by SSS.

When the equal angles are opposite the smaller of the two sides, the triangles need not be congruent.

An example: $\displaystyle ABC$ is isosceles, $\displaystyle {AB}={AC}$, and $\displaystyle D$ is a point of the base $\displaystyle BC$ that is not its midpoint.

• The triangles $\displaystyle ABD$ and $\displaystyle ACD$: $\displaystyle {AB}={AC}$, $\displaystyle AD$ is common, and $\displaystyle {\angle B}={\angle C}$.
• The equal angles $\displaystyle B$ and $\displaystyle C$ are opposite $\displaystyle AD$ – the smaller side, since $\displaystyle AD<AB$.
• The triangles are not congruent: $\displaystyle BD\neq DC$.

What it does give: two sides and the angle opposite the smaller of them – either the triangles are congruent, or the angles opposite the larger sides add up to $\displaystyle 180^{\circ}$. In the example, $\displaystyle {\angle ADB+\angle ADC}={180^{\circ}}$.

In the quadrilateral $\displaystyle ABCD$, $\displaystyle {AB}={CD}$ and $\displaystyle {\angle B}={\angle D}$.
The diagonal $\displaystyle AC$ is longer than $\displaystyle AB$.
Prove: $\displaystyle {BC}={AD}$.

Two sides of a triangle are $\displaystyle 8$ and $\displaystyle 5$, and the angle opposite the side $\displaystyle 5$ is $\displaystyle 30^{\circ}$.

1. Is the angle opposite the larger side or the smaller one?
2. Draw: a ray from $\displaystyle B$ at $\displaystyle 30^{\circ}$ to $\displaystyle {BC}={8}$, and a circle of radius $\displaystyle 5$ centred at $\displaystyle C$. How many triangles fit the data?
3. What can you say about the angles opposite the side $\displaystyle 8$ in the triangles you found?

Theorem: a triangle in which the sum of the squares of two sides equals the square of the third side is right-angled – the right angle opposite the third side.

Given: in the triangle $\displaystyle ABC$, $\displaystyle {BC}={a}$, $\displaystyle {AC}={b}$, $\displaystyle {AB}={c}$ and $\displaystyle {a^{2}+b^{2}}={c^{2}}$.
To prove: $\displaystyle {\angle C}={90^{\circ}}$.

The proof below builds a right-angled triangle with the same legs and compares:

• Build a triangle $\displaystyle DEF$ with $\displaystyle {\angle F}={90^{\circ}}$, $\displaystyle {EF}={a}$ and $\displaystyle {DF}={b}$.
• By Pythagoras’ theorem, $\displaystyle {DE^{2}}={a^{2}+b^{2}}={c^{2}}$, so $\displaystyle {DE}={c}$.
• So $\displaystyle \triangle ABC\cong\triangle DEF$ (SSS), and $\displaystyle {\angle C}={\angle F}={90^{\circ}}$.

The sides of a triangle are $\displaystyle 7$, $\displaystyle 24$ and $\displaystyle 25$.

Is the triangle right-angled? Which angle is right?
And a triangle with the sides $\displaystyle 6$, $\displaystyle 8$ and $\displaystyle 11$?

The vertices of the triangle $\displaystyle ABC$ are $\displaystyle A(1,1)$, $\displaystyle B(5,3)$ and $\displaystyle C(3,7)$.

Find the squares of the three sides.

What kind of triangle is $\displaystyle ABC$? Explain by the theorems of this lesson.