Geometric False Result Based on SSA Congruence Criterion for Triangles

Question

I'm writing an article to warn students not to use SSA (Side-Side-Angle) criterion to prove congruence between two triangles. I'd like to add to the article an example of a geometric fallacy based on SSA. Do you know any such fallacy?

Answer 1

Here is an example with $|AC|=1,|BC|=\sqrt{2},\text{ and }\angle ABC=30^\circ$

Alternate version of same diagram. Obviously the two triangles are not congruent even though they share two sides and an angle.

Answer 2

I tried create a fallacy. Maybe it is not totally right or it is not exactly what you want. But, maybe, it give you some idea.

Look the picture below.

By construction, $$\text{diameter of the blue circle}>\text{length of the black segment}+\text{length of the orange segment}$$

Furthermore:

• The red angles have tha same measure (by symmetry).

• The blue line segments have the same length (because they are radius of the same circle).

• The green line segments has the same length (because the green circles have the same radius).

It follows from the SSA criterion that the black and orange line segments have the same length. So,

\begin{align} \text{diameter of the blue circle}&>\text{length of the black segment}+\text{length of the orange segment}\\ &=\text{length of the orange segment}+\text{length of the orange segment}\\ &=2\times \text{radius of the blue circle} \end{align}

and thus, the formula $$\text{diameter}=2\times\text{radius}$$ is not valid in general.

Answer 3

Here's a fallacy relying on SSA.

Every triangle is right-angled.

Proof. Draw any circle $\Omega_1$, and pick any three points $A, B$ and $C$ on it. (As every triangle has a circumcircle, every triangle can be picked this way.)

Draw the circle with centre $A$ through $C$. The circles cross at $C$. If they do not cross at any other point, then only one point on $\Omega_1$ is the same distance from $A$ as $C$ is, so $AC$ is a diameter, angle $CBA$ is a right angle, and triangle $ABC$ is right-angled at $B$.

So suppose that the circles cross at another point; call it $D$. Draw $AD$ and $BD$.

There are two cases to consider.

Case 1. $C$ and $D$ are on the same side of $AB$. Consider triangles $ABC$ and $ABD$. Side $AB$ is common. Sides $AC$ and $AD$ are equal by construction. Angles $ACB$ and $ADB$ are in the same segment and are equal by Eucl. III Prop. 21. Therefore they are congruent, so angles $BAC$ and $BAD$ are equal, meaning that $D$ is the same point as $C$, so, by the argument above, triangle $ABC$ is right-angled at $B$.

Case 2. $C$ and $D$ are on opposite sides of $AB$. Then $ADBC$ is a cyclic quadrilateral. Consider triangles $ABC$ and $ADB$. The segments on $CA$ and $AD$ are congruent, so, by Eucl. III Prop. 26, angles $CBA$ and $ABD$ are equal. Side $AB$ is common. Sides $AC$ and $AD$ are equal by construction. Therefore triangles $ABC$ and $ADB$ are congruent, in opposite senses, reflected in $AB$. So angles $ACB$ and $BDA$ are equal. But they are opposite angles in the cyclic quadrilateral, so they sum to $180^\circ$, so they are right angles, and triangle $ABC$ is right-angled at $C$.

The flaws in the fallacy.

Each case inferred congruence from SSA, and this inference was invalid.

Case 1. $C$ and $D$ are indeed different points unless angle $CBA$ really is a right angle.

Case 2. The statement that angles $ACB$ and $BDA$ sum to $180^\circ$ is valid. The further inference that they are right angles is invalid.