CHAPTER 2

CONCURRENCY

2.1 Perpendicular Bisectors

This chapter is concerned with concurrent lines associated with a triangle. A family of lines is concurrent at a point P if all members of the family pass through P.

In preparation, we need a few additional facts about parallel lines.

Theorem 2.1.1. Let l₁ and l₂ be parallel lines, and suppose that m₁ and m₂ are lines with m₁ ⊥ l₁ and m₂ ⊥ l₂. Then m₁ and m₂ are parallel.

Proof. Let P be the point l₁ ∩ m₁ and let Q be the point l₂ ∩ m₂. By the parallel postulate, l₁ is the only line through P parallel to l₂, and so m₁ is not parallel to l₂ and consequently must meet l₂ at some point S. In other words, m₁ is a transversal for the parallel lines l₁ and l₂. Since the sum of the adjacent interior angles is 180°, it follows that l₂ must be perpendicular to m₁.

But l₂ is also perpendicular to m₂ and so m₁ and m₂ must be parallel.

In the above proof, if we interchange l₁ and m₁ and l₂ and m₂, we have:

Corollary 2.1.2. Suppose that l₁ ⊥ m₁ and l₂ ⊥ m₂. Then l₁ and l₂ are parallel if and only if m₁ and m₂ are parallel.

One of the consequences of this corollary is that if two line segments intersect, then their perpendicular bisectors must also intersect. This fact is crucial in the following theorem, the proof of which also uses the fact that the right bisector of a segment can be characterized as being the set of all points that are equidistant from the endpoints of the segment.

Theorem 2.1.3. The perpendicular bisectors of the sides of a triangle are concurrent.

Proof. According to the comments preceding the theorem, the perpendicular bisectors of AB and AC meet at some point O, as shown in the figure.

It is enough to show that O lies on the perpendicular bisector of BC, We have

OB = OA (since O is on the perpendicular bisector of AB),

OA = OC (since O is on the perpendicular bisector of AC).

Therefore OB = OC, which means that O is on the perpendicular bisector of BC.

The proof shows that the point O is equidistant from the three vertices, so with center O we can circumscribe a circle about the triangle. The circle is called the circumcircle, and the point O is called the circumcenter of the triangle.

Exercise 2.1.4. In the figure above, the circumcenter is interior to the triangle. In what sort of a triangle is the circumcenter

2.2 Angle Bisectors

Theorem 2.2.1. For a circle C(O, r):

Theorem 2.2.2. The internal bisectors of the angles of a triangle are concurrent.

Proof. In ΔABC, let the internal bisectors of ∠B and ∠C meet at I, as shown above.

We will show that I lies on the internal bisector of ∠A. We have

since I lies on the internal bisector of ∠B, and

since I lies on the internal bisector of ∠C.

Hence, d(I, AB) = d(I, AC), which means that I is on the internal bisector of ∠A.

The proof of Theorem 2.2.2 shows that if we drop the perpendiculars from I to the three sides of the triangle, the lengths of those perpendiculars will all be the same, say r. Then the circle C(I, r) will be tangent to all three sides by Theorem 2.2.1. That is, I is the center of a circle inscribed in the triangle. This inscribed circle is called the incircle of the triangle, and I is called the incenter of the triangle.

We will next prove a theorem about the external angle bisectors. First, we need to show that two external angle bisectors can never be parallel.

Proposition 2.2.3. Each pair of external angle bisectors of a triangle intersect.

Proof. Referring to the diagram,

Adding and rearranging:

Since γ ∠ 180, it follows that θ + ϕ ∠ 180, so the external angle bisectors cannot be parallel.

Theorem 2.2.4. The external bisectors of two of the angles of a triangle and the internal bisector of the third angle are concurrent.

Proof. Let the two external bisectors meet at E, as shown in the figure. It is enough to show that E lies on the internal bisector of ∠B.

Now,

since E lies on the external bisector of ∠A, and

since E lies on the external bisector of ∠C. Hence, d(E, AB) = d(E, BC), which means that E is on the internal bisector of ∠B.

The point of concurrency E is called an excenter of the triangle. Since E is equidistant from all three sides (some extended) of the triangle, we can draw an excircle tangent to these sides. Every triangle has three excenters and three excircles.

2.3 Altitudes

A line passing through a vertex of a triangle perpendicular to the opposite side is called an altitude of the triangle.

To prove that the altitudes of a triangle are concurrent, we need some facts about parallelograms.

A parallelogram is a quadrilateral whose opposite sides are parallel. A parallelogram whose sides are equal in length is called a rhombus. Squares and rectangles are special types of parallelograms.

Theorem 2.3.1. In a parallelogram:

Proof. We will prove (1) and (2). Given parallelogram ABCD, as in the figure below, we will show that ΔABC = ΔCDA.

Diagonal AC is a transversal for parallel lines AB and CD, and so ∠BAC and ∠DCA are opposite interior angles for the parallel lines. So we have

Treating AC as a transversal for AD and CB, we have

Since AC is common to ΔBAC and ΔDCA, ASA implies that they are congruent. Consequently, AB = CD and BC = DA, proving (1).

Also, ∠BAD = α + β = ∠BCD and ∠ADC = ∠ABC because triangles ADC and ABC are congruent, proving (2).

Exercise 2.3.2. Prove statement (3) of Theorem 2.3.1.

Theorem 2.3.3. If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

Exercise 2.3.4. Prove Theorem 2.3.3.

Theorem 2.3.3 tells us that statement (3) of Theorem 2.3.1 will guarantee that the quadrilateral is a parallelogram. However, neither statement (1) nor statement (2) of Theorem 2.3.1 is by itself enough to guarantee that a quadrilateral is a parallelogram. For example, a nonsimple quadrilateral whose opposite sides are congruent is not a parallelogram. However, if the quadrilateral is a simple polygon, then either statement (1) or (2) is sufficient. As well, there is another useful condition that can help determine if a simple quadrilateral is a parallelogram:

Theorem 2.3.5. A simple quadrilateral is a parallelogram if any of the following statements are true:

Proof. We will justify case (1), leaving the others to the reader.

Suppose that ABCD is the quadrilateral, as in the figure on the right. Since ABCD is simple, we may suppose that the diagonal AC is interior to the quadrilateral. Since the opposite sides of the quadrilateral are congruent, the SSS congruency condition implies that ΔABC = ΔCDA.

This in turn implies that the alternate interior angles ∠BAC and ∠DCA are congruent and also that the alternate interior angles ∠BCA and ∠DAC are congruent. The fact that the edges are parallel now follows from the well-known facts about parallel lines.

The next theorem uses a clever trick. It embeds the given triangle in a larger one in such a way that the altitudes of the given triangle are right bisectors of the sides of the larger one.

Theorem 2.3.6. The altitudes of a triangle are concurrent.

Proof. Given ΔABC, we embed it in a larger triangle ΔDEF by drawing lines through the vertices of ΔABC that are parallel to the opposite sides so that ABCE, ACBF, and CABD are parallelograms, as in the diagram on the right. Clearly, A, B, and C are midpoints of the sides of ΔDEF, and an altitude of ΔABC is a perpendicular bisector of a side of ΔDEF. However, since the perpendicular bisectors of ΔDEF are concurrent, so are the altitudes of ΔABC.

The point of concurrency of the altitudes is called the orthocenter of the triangle, and it is usually denoted by the letter H. In the proof, the orthocenter of ΔABC is the circumcenter of ΔDEF.

Note that the orthocenter can lie outside the triangle (for obtuse-angled triangles) or on the triangle (for right-angled triangles). The same proof also works for these types of triangles.

Exercise 2.3.7. The figure above shows a triangle whose orthocenter is interior to the triangle. Give examples of triangles where:

2.4 Medians

A median of a triangle is a line passing through a vertex and the midpoint of the opposite side.

Exercise 2.4.1. Show that in an equilateral triangle ABC the following are all the same:

The next theorem, which is useful on many occasions, is also proved by using the properties of a parallelogram.

Theorem 2.4.2. (The Midline Theorem)

If P and Q are the respective midpoints of sides AB and AC of triangle ABC, then PQ is parallel to BC and PQ = BC/2.

Proof. First, extend PQ to T so that PQ = QT, as in the figure. Then ATCP is a parallelogram, because the diagonals bisect each other. This means that TC is parallel to and congruent to AP. Since P is the midpoint of AB, it follows that TC is parallel to and congruent to BP, and so TCBP is also a parallelogram. From this we can conclude that PT is parallel to and congruent to BC; that is, PQ is parallel to BC and half the length of BC.

Lemma 2.4.3. Any two medians of a triangle trisect each other at their point of intersection.

Proof. Let the medians BE and CF intersect at G, as shown in the figure. Draw FE. By the Midline Theorem, FE is parallel to BC and half its length.

Let P be the midpoint of GB and let Q be the midpoint of GC. Then, again by the Midline Theorem, PQ is parallel to BC and half its length. Since FE and PQ are parallel and equal in length, EFPQ is a parallelogram, and so the diagonals of EFPQ bisect each other. It follows that FG = GQ = QC and EG = GP = PB, which proves the lemma.

There are two different points that trisect a given line segment. The point of intersection of the two medians is the trisection point of each that is farthest from the vertex.

Theorem 2.4.4. The medians of a triangle are concurrent.

Proof. Let the medians be AD, BE, and CF. Then BE and CF meet at a point G for which EG/EB = 1/3. Also, AD and BE meet at a point G′ for which EG′/EB = 1/3. Since both G and G′ are between B and E, we must have G = G′, which means that the three medians are concurrent.

The point of concurrency of the three medians is called the centroid. The centroid always lies inside the triangle. A thin triangular plate can be balanced at its centroid on the point of a needle, so physically the centroid corresponds to the center of gravity.

The partial converses of the Midline Theorem are useful:

Theorem 2.4.5. Let P be the midpoint of side AB of triangle ABC, and let Q be a point on AC such that PQ is parallel to BC. Then Q is the midpoint of AC.

Proof. Let Q′ be the midpoint of AC, then PQ′ || BC. Since there is only one line through P parallel to BC, the lines PQ and PQ′ must be the same, so the points Q and Q′ are also the same.

2.5 Construction Problems

The two basic construction problems associated with parallel lines are constructing a parallelogram given two of its adjacent edges (that is, completing a parallelogram) and constructing a line through a given point parallel to a given line. Once we have solved the first problem, the second one is straightforward.

Example 2.5.1. To construct a parallelogram given two adjacent edges.

Solution. Given edges AB and AC, with center B and radius AC, draw an arc. With center C and radius AB, draw a second arc cutting the first at D on the same side of AC as B. Then ABDC is the desired parallelogram.

Example 2.5.2. To construct a line parallel to a given line through a point not on the line.

Solution. With center P, draw an arc cutting the given line at Q and S. With the same radius and center S, draw a second arc. With center P and radius QS, draw a third arc cutting the second at T. Then PT is the desired line, because PQST is a parallelogram.

Example 2.5.3. Given a circle with center O and a point P outside the circle, construct the lines through P tangent to the circle.

Solution. Draw the line OP and construct the right bisector of OP, obtaining the midpoint M of OP. With center M and radius MP, draw a circle cutting the given circle at S and T. Then PS and PT are the desired tangent lines.

Note that angles OSP and OTP are angles in a semicircle, so by Thales’ Theorem both are right angles. Thus, PS and PT are perpendicular to radii OS and OT, respectively, and so must be tangent lines by Theorem 2.2.1.

Thales’ Locus

Given a segment AB and an angle θ, the set of all points P such that ∠APB = θ forms the union of two arcs of a circle, which we shall call Thales’ Locus, shown in the figure on the following page.

To help us construct Thales’ Locus, we need the following useful theorem.

Theorem 2.5.4. Let AB be a chord of the circle C(0,r) and let P be a point on the line tangent to the circle, as shown in the diagram below. Then ∠AOB = 2∠PAB.

Proof. Referring to the diagram, drop the perpendicular OM from O to AB. Then ∠AOB = 2∠AOM = 2α. Now, θ = 90 − ϕ = α, and it follows that ∠AOB = 2∠PAB.

Note. In a similar way, it can be shown that the reflex angle AOB is twice the size of ∠QAB.

Example 2.5.5. Given a segment AB and an angle θ < 90°, construct Thales’ Locus for the given data:

Data:

Solution. Here is one method of doing this construction.

By Theorem 2.5.4, ∠AOB = 2θ, so ∠AXB at the circumference is of size θ

Example 2.5.6. Construct a triangle ABC given the size θ of ∠A, the vertices B and C, and the length of the altitude h from A.

Solution. We give a quick outline of the solution.

Then ABC is the desired triangle.

2.6 Solutions to the Exercises

Solution to Exercise 2.1.4

In an acute-angled triangle, the circumcenter is always interior to the triangle.

Solution to Exercise 2.3.2

In triangles ABP and CDP we have

because AB || CD, and

by statement (1) of Theorem 2.3.1.

So, ΔABP ≡ ΔCDP by ASA, and it follows that AP = CP and BP = DP.

Solution to Exercise 2.3.4

In triangles ABP and CDP we have

since they are vertically opposite angles, and

So triangles ABP and CDP are congruent, and thus ∠PAB = ∠PCD, which implies that AB || CD.

Similarly, AD || CB, which shows that ABCD is a parallelogram.

Solution to Exercise 2.3.7

In an acute-angled triangle, the orthocenter is always interior to the triangle.

Solution to Exercise 2.4.1

We will show that the line through the vertex A and the midpoint D of BC is simultaneously the perpendicular bisector of BC, the bisector of ∠A, the altitude from A, and the median from A.

2.7 Problems