KindiakMath

Matchmaking Triangles

This is our motivating question for today:

What is the size of an angle in a triangle whose side lengths are equal to each other?

You might laugh at this question and reply: it is obviously

180^\circ \div 3 = 60^\circ.

And your final answer is right—but how did you know that all three angles are equal to each other?

To answer this question properly, we need to properly discuss congruent triangles, from which we can discuss isosceles triangles, of which equilateral triangles are a special case.

Definition 1. Consider the two triangles in the diagram below.

We say that \Delta ABC is congruent to \Delta A'B'C', denoted \Delta ABC \equiv \Delta A'B'C', if their corresponding side lengths and angles equal each other:

a = a', \quad b = b',\quad c = c',\quad \alpha = \alpha', \quad \beta = \beta',\quad \gamma = \gamma'.

In ordered tuple notation:

(a,b,c,\alpha,\beta,\gamma) = (a',b',c',\alpha',\beta',\gamma').

The topic of congruent triangles is commonly presented as a collection of “cookbook recipes” to determine if two triangles are effectively the same.

But how do we know that these recipes actually work? It turns out that we have the necessary “meta-recipes” to construct these recipes, and that’s what we shall do.

Theorem 1 (RHS Criterion). Let \Delta ABC, \Delta PQR be the following two right-angled triangles with \angle BCA = \angle QRP = 90^\circ:

Then \Delta ABC \equiv \Delta PQR if and only if

c = r\quad \text{and} \quad (a= p \quad \text{or} \quad b=q).

Proof. In the direction (\Rightarrow), we obtain all of the equalities by Definition 1.

For the direction (\Leftarrow), suppose

\displaystyle \angle BCA = \angle QRP = 90^\circ,\quad c = r, \quad a=p.

We will deal with the case b = q later. Observe that exactly one of the following hold:

\angle ABC = \angle PQR, \quad \angle ABC < \angle PQR, \quad \angle ABC > \angle QPR.

We claim that the remaining two cases are impossible.

Suppose \angle ABC < \angle PQR. Then we can overlay the triangles on their edges BC = QR as follows:

Hence, AP > 0 so that b < q. However, by Pythagoras’ theorem,

b = \sqrt{c^2 - a^2} = \sqrt{r^2 - p^2} = q,

a contradiction, ruling out the case \angle ABC < \angle PQR.

The case \angle ABC > \angle QPR can also be ruled out by relabelling (A,B,C) with (P,Q,R) and vice versa, and using the same reasons above (i.e. a symmetric argument).

Therefore, \angle ABC = \angle PQR.

Finally, since angles in a triangle sum to 180^\circ,

\begin{aligned} \angle BAC &= 180^\circ - \angle ABC - \angle BCA \\ &= 180^\circ - \angle PQR - \angle QRP = \angle QPR. \end{aligned}

Since all corresponding sides and angles equal each other, \Delta ABC \equiv \Delta PQR.

For the case b = q, we could use a symmetric argument, or take the following even shorter approach: by Pythagoras’ theorem,

a^2 = c^2 - b^2 = r^2 - q^2 = p^2,

so that a > 0 and p > 0 implies a = p, therefore

\displaystyle \angle BCA = \angle QRP = 90^\circ,\quad c = r, \quad a=p,

which, as shown just now, implies \Delta ABC \equiv \Delta PQR.

The RHS Criterion only describes congruence between right-angled triangles. Yet, more is true. Since right angles open our discussion on general angles, right-angled triangles also open our exploration of general triangles.

Theorem 2 (SAS Criterion). Let \Delta ABC, \Delta PQR be the following two triangles:

Then \Delta ABC \equiv \Delta PQR if and only if

b = q, \quad c = r,\quad \angle CAB = \angle RPQ.

Proof. The direction (\Rightarrow) is trivial.

For the direction (\Leftarrow), the hypothesis tells us that

AC = b = q = PR.

Construct the perpendicular heights (i.e. the altitudes) below.

Using rotation, we can assume AC \parallel PR. By construction,

\Delta ABD, \quad \Delta BCD, \quad \Delta PQS, \quad \Delta QRS

are all right-angled triangles. Our line of attack is as follows:

  • Show that BD = QS.
  • Deduce that \Delta ABD \equiv \Delta PQS using the RHS Criterion.
  • Similarly, \Delta BCD \equiv \Delta QRS.
  • Conclude that \Delta ABC \equiv \Delta PQR.

The first point is the most challenging.

Since AC \perp BD and AC \parallel PR and PR \perp QS, we must have BD \parallel QS. We already have AB = PQ by hypothesis, so it remains to prove BD = QS.

If BD < QS, then aligning B with Q, we can draw the diagram below.

By Pythagoras’ theorem,

c^2 = \sqrt{{AD}^2 + {BD}^2} < \sqrt{{PS}^2 + {QS}^2} = r^2.

Since c > 0 and r > 0, we have c < r, a contradiction. Similarly, we can rule out the case BD > QS. Therefore, BD = QS.

By the RHS Criterion, \Delta ABD \equiv \Delta PQS. In particular, BD = QS.

We next establish \Delta BCD \equiv \Delta QRS. By hypothesis, BC = QR. In particular,

\begin{aligned} CD &= BC - BD \\ &= QR - QS = RS. \end{aligned}

By Pythagoras’ theorem,

\begin{aligned} BC^2 &= {BD}^2 + {CD}^2 \\ &= {QS}^2 + {RS}^2 = QR^2. \end{aligned}

Since BC > 0 and QR > 0, we have BC = QR. By the RHS Criterion again, \Delta BCD \equiv \Delta QRS. In particular,

\begin{aligned} \angle BCA &= \angle BCD \\ &= \angle QRS = \angle QRP. \end{aligned}

Finally,

\begin{aligned} \angle ABC &= \angle ABD + \angle CBD \\ &= \angle PQS + \angle RQS = \angle PQR. \end{aligned}

Remark 1. This proof holds true in all three cases (i.e. \angle ABC and \angle PQR can be acute, right, or obtuse), so long as both of the other two angles are acute.

Once we have a congruence criterion test that involves both sides and angles, we can derive many other useful congruence criterion tests.

Theorem 3 (SSS Criterion). Let \Delta ABC, \Delta PQR be the following two triangles:

Then \Delta ABC \equiv \Delta PQR if and only if

a=p,\quad b = q, \quad c = r.

Proof. The direction (\Rightarrow) is trivial.

For the direction (\Leftarrow), AC = PR by hypothesis. Construct the altitudes once again.

If a=p and c= r, then by Pythagoras’ theorem, BD < QS implies

\begin{aligned} AC &= \sqrt{a^2 - d^2} + \sqrt{c^2 - d^2} \\ &> \sqrt{p^2 - s^2} + \sqrt{r^2 - s^2} = PR, \end{aligned}

so that AC > PR, a contradiction. Therefore, by a symmetric argument, we can conclude that BD = QS.

By the RHS Criterion, \Delta ABD \equiv \Delta PQS, so that

\angle CAB = \angle DAB = \angle SPQ = \angle RPQ.

By hypothesis, AB = PQ and AC = PR. Hence, by the SAS Criterion,

\Delta ABC \equiv \Delta PQR,

as required.

Example 1. Consider the triangle \Delta ABC below.

Prove that \theta = 90^\circ if and only if a^2 + b^2 = c^2. The direction (\Leftarrow) is called the converse of Pythagoras’ theorem.

Solution. The direction (\Rightarrow) is the already-proven vanilla Pythagoras’ theorem.

For the direction (\Leftarrow), suppose a^2 + b^2 = c^2. Construct a right-angled triangle \Delta PQR with base b and height a as follows:

By the vanilla Pythagoras’ theorem, a^2 + b^2 = d^2. By hypothesis, c^2 = d^2.

Since c > 0 and d > 0, we have c = d.

By the SSS Criterion, \Delta ABC \equiv \Delta PQR. In particular,

\theta = \angle BCA = QRP = 90^\circ.

Theorem 4 (ASA Criterion). Let \Delta ABC, \Delta PQR be the following two triangles:

Then \Delta ABC \equiv \Delta PQR if and only if

\angle BAC = \angle QPR, \quad b=q,\quad \angle BCA = \angle QRP.

Proof. It suffices to prove (\Leftarrow). By hypothesis, since angles in a triangle sum to 180^\circ,

\begin{aligned} \angle ABC &= 180^\circ - \angle BAC - \angle BCA \\ &= 180^\circ - \angle QPR - \angle QRP = \angle PQR. \end{aligned}

Similar to the proof of Theorem 2, we claim that AB = PQ. Superimpose \Delta ABC onto \Delta PQR such that A = P and C = R.

Since the corresponding angles \angle ACB = \angle PRQ and AC = PR, we have BC \parallel QR. Since C = R, BC passes through R. By Playfair’s axiom, B must lie on QR. Using a symmetric argument, B must lie on PQ.

Therefore, B lies on the intersection between the line PQ and the line QR. Since there exists one and only one intersection point between the two lines, namely Q, we must have B = Q, so that BC = QR.

By hypothesis, \angle BCA = \angle QRP and AC = PR. Hence, by the SAS Criterion, \Delta ABC \equiv \Delta PQR, as required.

Theorem 5 (AAS Criterion). Let \Delta ABC, \Delta PQR be the following two triangles:

Then \Delta ABC \equiv \Delta PQR if and only if

\angle ABC = \angle PQR, \quad \angle BCA = \angle QRP,\quad AC = PR.

Proof. Since angles sum to 180^\circ,

\begin{aligned} \angle CAB &= 180^\circ - \angle ABC - \angle BCA \\ &= 180^\circ - \angle PQR - \angle QRP = \angle RPQ. \end{aligned}

Therefore, by the ASA Criterion, \Delta ABC \equiv \Delta PQR, as required.

Definition 2. A triangle is said to be isosceles if it has two sides with equal length.

Example 2. Consider the triangle \Delta ABC below with base angles \angle BAC and \angle BCA.

Prove that AB = BC if and only if \angle BAC = \angle BCA. In this case, we say that the base angles of an isosceles triangle are equal.

Solution. Our proof boils down to the analysis of \Delta ABC with its “rearranged” self \Delta CBA. Clearly \angle ABC = \angle CBA, since they denote the same angle.

We prove in two directions. For the direction (\Rightarrow), suppose AB = BC. Almost trivially, AB = CB and BC = BA. By the SAS Criterion, \Delta ABC \equiv \Delta CBA. In particular, \angle BAC = \angle BCA.

The direction (\Leftarrow) is proven similarly. Suppose \angle BAC = \angle BCA. Then by relabelling, \angle ACB = \angle CAB. Furthermore, AC = CA, since they denote the same side. By the ASA Criterion, \Delta ABC \equiv \Delta CBA. In particular, AB = BA = BC, as required.

Definition 3. A triangle is said to be equilateral if all three sides have the same length.

Example 3. Determine the size of any interior angle in an equilateral triangle.

Proof. Denote the equilateral triangle by \Delta ABC, and denote \angle CAB = \theta.

Since CA = CB, by Example 2,

\angle CBA = \angle CAB = \theta.

Since AC = AB, by Example 2 again,

\angle ACB = \angle ABC = \angle CBA = \theta.

Since the angles in a triangle sum to 180^\circ,

\begin{aligned} \angle CAB + \angle CBA + \angle ACB &= 180^\circ \\ \theta + \theta + \theta &= 180^\circ \\ 3\theta &= 180^\circ \\ \theta &= 60^\circ. \end{aligned}

Therefore,

\angle CBA = \angle CAB = \angle ACB = 60^\circ.

That is, each angle in an equilateral triangle has a size of 60^\circ.

If the geometry on a triangle is fascinating, the geometry on a circle is even more astounding! We will visit the geometry of circles in the next post.

—Joel Kindiak, 14 Oct 25, 1920H

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