1 . The line passes through the center of the circle.
2. The line passes through the midpoint of the chord.
3. The line is perpendicular to the chord.
If any two of these statements are true, then all three are true.
Fact 8.2. 14 A tangent line to a circle is perpendicular to the radius at the point of tangency; conversely, the perpendicular to a tangent line at the point of tangency will pass through the center of the circle.
8.2.15 Let A be a point outside a circle. Draw lines AX , AY tangent to the circle at X
and Y . Prove that AX = AY .
3Notice that there is an ambiguity about arcs. Does arc Be mean the path along the circumference going clockwise or counterclockwise? If it is the former, the central angle will be more than 1 800 • We will use the convention that arcs are read counterclockwise.
4We are indebted to Andy Liu, who used this clever "three for two sale" formulation in [28].
The following lemma is perhaps the first interesting and unexpected result proven in high-school geometry.
Example 8.2.16 The Inscribed Angle Theorem. The measure of an arc is twice the measure of the inscribed angle that subtends it. (In other words, for any arc, the angle subtended at the center is twice the angle subtended at the circle.)
This is truly remarkable. It says that no matter where we place the point A on the circumference of the circle r below, the measure of the angle CAB will be fixed (and moreover, equal to half of the central angle COB).
Solution : Let (J = LCOB. We wish to show that LCAB = (J /2. In order to do this, we need to gather as much angular information as we can. It certainly makes sense to draw in one auxiliary object, namely, line segment AO, since this adds two new triangles that have a vertex at A. And moreover, these triangles are isosceles, since
AO, BO, and CO are radii of circle r.
�c r B
Let f3 = LAOB. Our strategy is simple: we will compute LCAB by finding LCAO and LBAO. And these are easy to find, since by Fact 8.2. 1 1 , we have
and
LBAO = 90 - f3 -2 '
LCAO = 90 _ 360 -2 f3 -(J = Ii 2 2 + � -90.
Adding these, we conclude that
LCAB = LCAO + LBAO = �. •
That wasn 't too hard ! A single, fairly obvious auxiliary line, followed by as much angular information as possible, led almost inexorably to a solution. This method is called angle chasing. The idea is to compute as many angles as possible, keeping the number of variables down. In the above example, we used two variables, one of which cancelled at the end.
Angle chasing is easy, fun, and powerful. Unfortunately, many students use it all the time, perhaps with trigonometry and algebra, to the exclusion of other, more elegant tactics. A key goal of this chapter is to teach you how to get the most from angle chasing, and when you should transcend it.
Here are a few simple and useful corollaries of the inscribed angle theorem.
266 CHAPTE R 8 G EO M ETRY FOR A M E R I CANS
8.2.17 Inscribed Right Triangles. Let AB be the diameter of a circle, and let C be an arbitrary point on the circle. Then LBCA = 90°. Conversely, if AB is a diameter and
LBCA = 90°, then C must lie on the circle. (See left figure below.)
I!---"T B
A c
B
8.2.18 Cyclic Quadrilaterals. The quadrilateral ABCD is cyclic if its vertices lie on a circle. The points A, B, C, D are called concyclic. (See right figure above.)
(a) A quadrilateral is cyclic if and only if a pair of opposite angles are supplementary.
(b) Points A , B, C, D are concyclic if and only if LACB = LADB.
With cyclic quadrilaterals you automatically get circles and pairs of equal angles "for free." This extra structure often provides useful information. Get in the habit of locat
ing (or creating) cyclic quadrilaterals.
Circles and Triangles
One can spend a lifetime exploring the interplay between circles and triangles. First we shall look at the inscribed and circumscribed circles of a triangle.
In the diagram below, triangle ABC is inscribed in circle n , since each vertex of the triangle lies on this circle. We call n the circumscribed circle or circumcircle of triangle ABC. The center of n , or circumcenter, is 0, and the length OA is the circumradius.
Likewise, circle 12 is inscribed in triangle ABC because it is tangent to each side of the triangle. We call it the inscribed circle or incircle. The center of 12, the incenter, is I, one point of tangency is D, and thus the length of I D is the inradius.
It is not immediately obvious that an arbitrary triangle even has an inscribed or circumscribed circle. Nor is it obvious that if such circles exist, that they be unique.
Let's explore these questions, starting with the circumscribed circle.
8.2. 19 Suppose a triangle possesses a circumscribed circle. Then the center of this circle is the intersection point of the perpendicular bisectors of the sides of the triangle.
(Consequently, if there is a circumscribed circle, then it is unique.)
8.2.20 Given any triangle, consider the intersection of any two perpendicular bisectors of its sides. This intersection point will be equidistant from the three vertices; hence will be the center of the circumscribed circle.
8.2.21 Thus the circumcenter is the intersection point of all three perpendicular bisec
tors. (So we get, "for free," the interesting fact that the three perpendicular bisectors intersect at a single point.)
Likewise, there is a (unique) inscribed circle whose center is the intersection point of three special lines. Again, our strategy is to first assume that the inscribed circle exists, and explore its properties. The tricky part, as above, is uniqueness.
8.2.22 Every triangle has a unique inscribed circle. The center is the intersection point of the three angle bisectors of the triangle. (Hint: Suppose the inscribed circle exists.
Show that its center is the intersection of two angle bisectors, using Fact 8.2. 14. Then show that the converse is true; i.e., the intersection of any two angle bisectors is the center of an inscribed circle. Finally, show that there cannot be more than one such circle, and hence all three bisectors meet in a single point.)
The existence of the circumcircle and incircle led to the wonderful facts that in any triangle, the three angle bisectors are concurrent (meet in a single point), and the three perpendicular bisectors of the sides are concurrent. There are many other
"natural" lines in a triangle. Which of them will be concurrent?
We will explore this is greater detail in Section 8.4, but here is a nice example that uses an ingenious auxiliary construction, and not much else.
Example 8.2.23 Show that, for any triangle, the three altitudes are concurrent. This point is called the orthocenter of the triangle.
Solution : An altitude is a line (not a line segment) that passes through a vertex and is perpendicular to the opposite side (extended, if necessary). Informally, the word refers to a line or a segment, or a segment length, depending on context. For example, in the figure below, the altitude from B to side AC is the actually the line BD, but it is not uncommon for the segment BD or its length (also written BD) to be called the altitude. The point D at which the altitude intersects the side is called the foot
of the altitude. In standard usage, the foot is not explicitly mentioned; for example,
"altitude BD" instead of the more precise "drop a perpendicular from vertex B to side
AC, intersecting it at D."
Notice that altitudes may not always lie inside the triangle. The altitude through C meets the extension of side AB at E, outside the triangle. Notice also that altitudes CE
268 CHAPTER 8 G EO M ETRY FOR A M E R I CANS
and BD intersect outside the triangle (remember, they are lines, not line segments).
�c A / /
/
�E
How do we show that the three altitudes meet in a single point? Well, what other lines are concurrent? The perpendicular bisectors are almost what we want, and they meet in a single point, the circumcenter. However, altitudes are perpendicular to opposite sides. In general, they do not bisect them. But is there any way to make them bisect something? Yes ! The ingenious trick is to make the point of bisection be the vertex, not the side.
In the figure below, we start with an arbitrary triangle ABC, and then draw lines through each vertex that are parallel to the opposite sides. These three parallel lines form a larger triangle, E F D, that contains ABC.
F
Let m be the altitude of triangle ABC that passes through B. Certainly, m is per
pendicular to AC, by definition. But since DE II AC, then m is also perpendicular to DE .
Notice that ADBC is a parallelogram, so AC = BD by Example 8.2.9. Likewise,
AC = BE . Thus B is the midpoint of DE , so m is the perpendicular bisector of DE.
By the same reasoning, the other two altitudes of triangle ABC are perpendicular bisectors of EF and F D. Consequently, these three altitudes meet in a point, namely
the circumcenter of triangle DEF ! •
Problems and Exercises
The problems below are elementary, in that no geometric facts other than those developed in this section are needed for their solution. Of course, "elementary" does not mean "easy." Some of them may be easier to do after you read the next section. But do try them all. The problems include many formulas and important ideas that will be used in later sections.
8.2.24 Let ABC be isosceles with vertex angle A, and I. Centroid
let E and D be points on AC and AB, respectively, so 2. Circumcenter that
AE = ED = DC =CB.
Find LA.
8.2.25 I n the previous problem, w e chose points on the equal sides of an isosceles triangle to create a
"chain" of four equal segments, including the base of the triangle. Can you generalize to n-segment chains?
8.2.26 Prove that the midpoints of the sides of an arbi
trary quadrilateral are the vertices of a parallelogram.
8.2.27 Let ABC be a right triangle, with right angle at C. Prove that the length of the median through C
is equal to half the length of the hypotenuse of ABC.
Thus if CE is the median, we have the nice fact that
AE = BE = CE.
8.2.28 Compass-and-Ruler Constructions. Undoubt
edly, you learned at least a few compass-and-ruler con
structions (also called Euclidean constructions). Re
view this important topic by finding the following con
structions (make sure that you can prove why they work). You may use a compass and an unmarked ruler.
No other tools allowed (besides a pencil).
(a) Given a line segment, find its midpoint.
(b) Given a line segment, draw its perpendicular bi
sector.
(c) Given a line e and a point P not on e, draw a line parallel to e that passes through P.
(d) Given a line e and a point P not on e, draw a line perpendicular to e that passes through P.
(e) Given an angle, draw its bisector.
(f) Given a circle, find its center.
(g) Given a circle and a point exterior to it, draw the tangent to the circle through the point.
(h) Given a line segment of length d, construct an equilateral triangle whose sides have length d.
(i) Given a triangle, locate the
3. Orthocenter 4. Incenter
U) Given a circle, with two given points P. Q in its interior, inscribe a right angle in this circle, such that one leg passes through P and one leg passes through Q. The construction may not be possi
ble, depending on the placement of P, Q.
(k) Given two circles, draw the lines tangent to them.
8.2.29 The triangle formed by joining the midpoints of the sides of a given triangle is called the medial tri
angle.
(a) Prove that the medial triangle and the original triangle have the same centroid. (For a much harder variation on this, see Problem 8 . 3 .4 1 .) (b) Prove that the orthocenter (intersection of alti
tudes) of the medial triangle is the circumcenter of the original triangle.
8.2.30 Let e, and e2 be parallel lines, and let OJ and
y be two circles lying between these lines so that P I is tangent to OJ, OJ is tangent to y, and y is tangent to P2 . Prove that the three points of tangency are collinear, i.e., lie on the same line.
8.2.3 1 (Leningrad Mathematical Olympiad 1 987) Al
titude CH and median BK are drawn in an acute trian
gle ABC, and it is known that BK = CH and LKBC = LHCB. Prove that triangle ABC is equilateral.
8.2.32 (Leningrad Mathematical Olympiad 1 988) Acute triangle ABC with LBAC = 30° is given. Alti
tudes BB, and CC, are drawn; B2 and C2 are the mid
points of AC and AB, respectively. Prove that segments
B,C2 and B2C, are perpendicular.
8.2.33 (Mathpath 2006 Qualifying Quiz) Consider the following "recipe" for folding paper to get equilateral triangles (see the figure below):
270 CHAPTER 8 G E O M ETRY FOR A M E R I CANS
I. Start with a long strip of paper, and visualize a crease for folding (thin line). The crease can have any angle.
2. Hold the top left comer and fold it down on this crease so that the comer is now below the bot
tom of the strip.
3. Unfold. You now actually have a crease (shown by thin line).
4. Now grab the right end of the strip and fold DOWN so that the top side of the strip is along this crease.
5. Unfold. You now have two creases.
6. Now grasp the right end and fold UP so that the bottom side of the strip is along the most re
cently created crease.
7. Unfold. You now have three creases.
8. Repeat steps 4--7. Your creases will now be equilateral triangles!
Comment on this procedure. Does it work? Does it almost work? Explain!
�
6) � 7)1::2/\::;ZZ::::::
1) I I I 2)f==l
4) 5)1 1\
8.2.34 (Bay Area Mathematical Olympiad 1 999) Let C be a circle in the xy-plane with center on the y-axis and passing through A = (O, a) and B = (O,b) with
° < a < b. Let P be any other point on the circle, let Q 8.3 Survival Geometry I I
Area
be the intersection of the line through P and A with the x-axis, and let 0 = (0,0). Prove that LBQP = LBOP.
8.2.35 (Canada 1 99 1 ) Let C be a circle and P a given point in the plane. Each line through P that intersects C determines a chord of C. Show that the midpoints of these chords lie on a circle.
8.2.36 (Bay Area Mathematical Olympiad 2000) Let
ABC be a triangle with D the midpoint of side AB, E
the midpoint of side BC, and F the midpoint of side AC. Let kl be the circle passing through points A, D,
and F; let k2 be the circle passing through points B, E,
and D; and l e t k3 b e the circle passing through C, F , and E. Prove that circles k I , k2 ' and k3 intersect i n a point.
8.2.37 Let ABC be a triangle with orthocenter H (re
call the definition of orthocenter in Example 8.2.23).
Consider the reflection of H with respect to each side of the triangle (labeled HI , H2 , H3 below).
• HI
B
Show that these three reflected points all lie on the cir
cumscribed circle of ABC.
We shall treat the notion of area intuitively, considering it to be "undefined," like points and lines. But area is not an "object," it is a/unction: a way to assign a non-negative number to each geometric object. Everyone who has ever cut construction paper has internalized the following axioms.
• Congruent figures have equal areas.
• If a figure is a union of non-overlapping parts, its area is the sum of the areas of its component parts.
• If a figure is a union of two overlapping parts, its area is the sum of the areas of the two parts, minus the area of the overlapping region. This is the geometric
version of the Principle of Inclusion-Exclusion (Section 6.3).
But how do we compute area? We need just one more axiom, which can serve to define area:
The area of a rectangle is the product of its base and height.
Long ago, you learned that the area of a triangle is "one-half base times height."
Let's rediscover this, using the axioms above. The first fact we need is that area is invariant with respect to shearing.
Example 8.3.1 Two parallelograms that share the same base and whose opposite sides lie on the same line have equal area.
Proof' In the diagram below, parallelograms ABCD and ABEF share the base AB with opposite sides lying on the same line (which of course is parallel to AB).
We use the notation [.] for area. Notice that
[ABCD] = [ABG] + [BCE] - [EGD] ' and
[ABEF] = [ABG] + [ADF] - [EGD] .
It is easy to check that triangles ADF and BCE are congruent (why?), and hence
[ADF] = [BCE] . Thus [ABCD] = [ABEF] . •
Since we know how to find the area of a rectangle, and since we can always "glue"
two copies of a triangle to form a parallelogram, we easily deduce the classic area formulas below, along with a very important corollary.
8.3.2 The area of a parallelogram is the product of the base and the height (where
"height" is the length of the perpendicular from a vertex opposite the base to the base).
8.3.3 The area of a triangle is one-half of the product of base and height.
8.3.4 If two triangles share a vertex, and the bases opposite this common vertex lie on the same line, then the ratio of the areas is equal to the ratio of the bases. For example, if BD = 4 and BC = 15 below, then
[ABD] = B [ABe] . 4
272 CHAPTER 8 G EO M ETRY FOR A M E R I CANS
~ B D C
We should point out that 8.3. 1 and 8.3.2 are intuitively obvious by visualizing a parallelogram as a "stack" of lines (like a deck of cards, but with infinitely many
"cards" of zero thickness): sliding the stack around, as long as it stays parallel to the base, shouldn't change the area. The area should equal the base length (which is the constant cross-sectional length) times the height. In the figure below, the two areas are equal.
Let's employ this idea-the shearing tool-to prove the most famous theorem in ele
mentary mathematics.
Example 8.3.5 Prove the Pythagorean Theorem, which states that the sum of the squares of the legs of a right triangle equals the square of the hypotenuse.
Proofã Let ABC be a right triangle with right angle at B. Then we wish to prove that
This is something new: an equation involving lengths multiplied by lengths. So far, only one geometrical concept allows us to think about such things, namely area. So we are naturally led to the most popular recasting of the Pythagorean Theorem: to show that the sum of the areas of the two small squares below is equal to the area of the largest square.
The beautiful crux idea is an auxiliary line m drawn through the right angle (B), per
pendicular to the hypotenuse AC. Line m divides the big square-on-the-hypotenuse into two rectangles, RSIA and RCHS. If we slide the line segment RS along m, these rectangles will shear into parallelograms, but their areas will not change.
In particular, we can slide RS so that R coincides with B. Meanwhile, we can play a similar game with the two smaller squares-on-the-legs. If we shear BAFG into a parallelogram by sliding segment FG straight down to m (so that G lies on m), the area will not change. Likewise, shear BDEC by sliding DE to the left. The result is below.
Now it's obvious-the two "leg" squares, and the two "hypotenuse" rectangles sheared into congruent parallelograms ! So of course the areas are equal ! _
8.3.6 OK, it's not quite obvious. Prove rigorously that the parallelograms are indeed congruent. You 'll need to do a little angle chasing to get congruent copies of the original triangle.
The Pythagorean Theorem has literally hundreds of different proofs. Perhaps the sim
plest, discovered in ancient India, uses nothing but "cutting and pasting."
Example 8.3.7 A "dissection" proof of the Pythagorean Theorem. The picture below should make it clear. The area of the first square is a2 + b2 plus four times the area of the right triangle. The area of the second square is c2 plus four times the area of the right triangle. Since the squares are congruent, their areas are equal, so we conclude
that a2 + b2 = c2. _