Chapter 7: Triangles and Circles Circumcircles Our main goal of this section is to show that for any triangle there exists a unique circle that contains the three vertices, A, B, and C. We say that this circle is circumscribed about and call it the circumcircle of . (We also say that is inscribed in the circle.)

The center of the circumcircle is called the circumcenter, and it will be denoted O. The radius of the circumcircle is called the circumradius, and it will be denoted R. The key step in the proof is the following lemma: Lemma. Given a line segment and a point P, the point P will be on the perpendicular bisector of if and only if P is equidistant from A and B (i.e., ). Proof. Let M be the midpoint of . We will show that is perpendicular to if and only if .

Since we are proving an if and only if statement, we have two statements to prove. First, we assume that . In this case by SSS. So and since the sum is , each must be . . Next, assume that is perpendicular to . In this case since we use SAS to conclude that . . So .

We now prove our main theorem. Theorem. Given any triangle , there exists a unique circle circumscribed about it. Proof. Let be the perpendicular bisector of and let be the perpendicular bisector of . Let O be the intersection point of and . Apply the lemma to see that, since O is on , OB = OC, and, since O is on , OA = OC.

Combining these two equations, we see that OA = OB = OC Combining these two equations, we see that OA = OB = OC. Hence, O is equidistant from the three points A, B, and C and there is a circle with center O containing these three points. Now suppose that is the center of a circle that passes through each of A, B, and C. Because and another application of the lemma, must lie on both and ,and hence . Therefore, only one circle contains A, B, and C and the circumscribed circle we have constructed is unique.

As a corollary of the proof we get the next result. Corollary. In any triangle the three perpendicular bisectors of the sides , , and meet in a point. We remark that this proof is constructive. Given we not only know that a circumcircle exists but we also have a technique to find O and we may draw the circle. Although O is the center of in the sense that it is the unique point equidistant from A, B, and C, we point out that O does not have to lie inside the triengle. O lies inside of if and only if is acute.

A Theorem of Brahmagupta When you first learned about areas you were told that the area of a rectangle was length times width. It might have occurred to you that the area of a triangle should be the product of the three sides, abc. Of course it isn’t, but in this section we will prove that the product abc is related to area. We start off with a consideration of the product of two sides of a triangle.

Theorem. Given , let AC = b, AB = c, let R be the circumradius, and let h = the length of the altitude from A to . Then bc = 2Rh. Proof. Let be the point such that is a diameter of the circumcircle and consider the triangles and in the Figure. The angles and each subtend the arc AC and are congruent, . Also, is inscribed in a semicircle, so it is a right angle.

Therefore, . By AA, is similar to . .So Or, by cross multiplication, . But AD = h, AB = c, AC = b, and the theorem follows. As an immediate consequence we get the following elegant theorem due to Brahmagupta. We use the letter K to denote the area of triangle . Theorem. abc = 4RK. Proof. By the previous theorem, bc = 2Rh. Now multiply both sides by a to get abc = 2Rah. But K = ah and the theorem follows.

Inscribed Circles In this section we will prove that given any triangle there is a unique circle inscribed in called the incircle of . A circle is said to be inscribed in a triangle if each of the three sides is tangent to the circle. The triangle is said to be circumscribed about the circle. The center of the incircle (inscribed circle) is called the incenter and is denoted I. The radius is called the inradius and is denoted r. In order to prove the existence and uniqueness of incircles, we need a lemma on angle bisectors.

Lemma. Let be any angle and let P be a point in the interior of Lemma. Let be any angle and let P be a point in the interior of . Then P is on the bisector of if and only if P is equidistant from the sides of .

Proof. Let and be the perpendiculars from P to the sides of Proof. Let and be the perpendiculars from P to the sides of . We need to show that if and only if . First assume that . Then by SSA for right triangles. Since and are corresponding parts, they must be congruent. Next, assume that . Then by SAA, and are congruent for they have two angles congruent and share a side. Hence . Theorem. Given any triangle , there is a unique circle inscribed in it.

Proof. Let be the bisector of and let be the bisector of Proof. Let be the bisector of and let be the bisector of . Let and intersect at the point I and apply the lemma: Since I is on the bisector of , the distance from I to equals the distance from I to . Similarly, since I is on the bisector of , the distance from I to equals the distance from I to .

Combining these, we see that I is equidistant from the three sides , , and . So there is a circle with center I that will be tangent to all three. To show that this circle is the unique incircle, suppose that P is the center of a circle D inscribed in . Drop segments from P to E and F, the points of tangency of the lines and , respectively, with D. We know from previous work that and ; moreover, PE = PF since both are radii of D. Applying our lemma, we see that P is on , the bisector of . Similarly, P is also on , which implies that P = I and that the first circle we constructed is unique.

As a corollary of the proof we get the next result. Corollary. Given any triangle , the bisectors of the three angles meet in one point. We point out that our proof here is also constructive. Given , we can construct I by finding the meeting point of the angle bisectors and we can use I to construct the incircle by dropping perpendiculars. One way in which I is better behaved than O is that I always lies inside the triangle. This fact has an interesting implication for area.

Theorem. If K = the area of , r = the inradius, and s = times the perimeter, then K = sr. Proof. Draw IA, IB, and IC as in the Figure. It is clear that the area of is the sum of the areas of , , and . Consider . The base is , which has length c, and the height is the distance from I to AB, which is r. So area . Likewise, has area and has area . By addition,

An Old Chestnut (The Steiner-Lehmus Theorem) Theorem 1. In let the bisector of meet at D and let the bisector of meet at E. If , then Theorem 2. In let the altitude of the side meet it at D and let the altitude of the side meet it at E. If , then

Theorem 2. In let the median of the side meet it at D and let the median of the side meet it at E. If , then Proof of Theorem 1. The proof will be by contradiction. We will assume that but that is larger than Since is greater than , , which is half of , is greater than half of . So we can find a point F strictly between A and D such that . Let G be the intersection point of and .

Now consider and. Each has for one angle. Also, , since each is Now consider and . Each has for one angle. Also, , since each is . Hence, by AA, and thus By hypothesis CE = BD, which forces . This implies that and, by the preceding equation, that Now that we have shown that we can get a contradiction, because we have another line of reasoning that will show the opposite. Consider . The angle , because we assumed that . But, in any triangle, a larger angle must be opposite a larger side. So This is a contradiction and so our initial assumption is impossible. Assuming will yield the same contradiction. Thus the theorem is proved.

Excribed Circles Before discussing escribed circles, it is worthwhile to define external bisectors of angles. Let be any angle with bisector .Extend one of the sides of the angle, say , in the other direction to form , the supplementary angle. Then the line which bisects this angle is called the external bisector of

External bisectors have two simple properties that will be important to us. Lemma. Let have internal bisector and external bisector . Then (1) is perpendicular to . (2) The points of are equidistant from and . Proof. (1) (2) The points of are on the bisector of and so they are equidistant from and .

Theorem. Let be any triangle Theorem. Let be any triangle . Then the external bisector of , the external bisector of , and the internal bisector of all meet in a point . Moreover, there is a circle with center tangent to the three lines , , and .

The circle we constructed in this manner is said to be an excribed circle for , the point is called an excenter, and the radius of the circle, which we denote , is called an exradius. The triangle has two more excircles, one touching and one touching . We denote their centers by and their radii as and , respectively. Our next theorem generalizes the area formula to the case of excircles. Theorem. Let K = the area of , = the radius of the excribed circle opposite , BC = a, and s = the semiperimeter. Then .

Proof. The area of = the area of + the area of the area of (See the Figure). Taking as the base, we see that has area , taking as base has area and taking as base, has area . Hence the area of

Corollary. If r is the inradius of and K is its area, then . Proof. By the theorem, . Likewise and . Also, since , . Hence by Heron’s formula. Corollary. Proof.

Euler’s Theorem In this section we will prove a celebrated theorem due to Euler that calculates the distance from the incenter I to the circumcenter O. Theorem. The proof will depend upon a two part lemma. In the circumcircle of pictured in the Figure we will denote by M the midpoint of arc .

Lemma. (1) M lies on the perpendicular bisector of and the bisector of . (2) There is a circle with center M that contains the points I, B, C, and . Proof. (1) Since , . But since M is equidistant from B and C, it must lie on the perpendicular bisector of . Also, since they subtend equal arcs. (2) First notice that bisects and that bisects , where D is a point on extended. Since is a straight angle, the angle is a right angle. Similarly, . is a right angle.

Hence, if we draw a circle with as diameter it will contain the points B and C. M must be the center of this circle because it lies on the intersection of a diameter with the perpendicular bisector of a chord. Proof of Euler’s Theorem. Extend to a chord of the circumcircle. The line extends to the chord . Hence The rest of the proof will consist of an examination of each side of the above equation.

We first consider the left-hand side We first consider the left-hand side. PI = OP – OI = R – OI and IQ = IO + OQ = (R+OI). Hence PI.IQ = (R-OI)(R+OI) = To evaluate the right-hand side we draw perpendicular to and extend to a diameter . We claim that . This is because and . Hence .

What can we say about these lengths What can we say about these lengths? By the lemma CM = IM, since they are radii of the same circle. is a diameter of the circumcircle and has length 2R.IZ is the distance from the incenter to a side of the triangle, so it is r. Making these three substitutions we see that , so AI . IM =2Rr. Now we substitute the results of each of the preceeding paragraphs into the equation to get or Corollary.

Center of Gravity In let the midpoints of sides , , and be , , and , respectively. The line segments , , and connecting the three vertices to the midpoints of the opposite sides are called the medians of . Our main theorem states that the three medians meet in a point G called the centroid of gravity of .

Lemma. Let be the midpoint of and be the midpoint of Lemma. Let be the midpoint of and be the midpoint of . Then is parallel to and is half as long. Proof. The triangles and are similar but the SAS theorem for similar triangles because , , and . Hence and , since they make equal corresponding angles with the transversal . Finally, Let us label the intersection point of and as G. We don’t yet know that G is also on . We first prove, as an intermediate step, that G is the trisection point of .

Lemma. Proof. Since is parallel to , and . since they are alternate interior angles. Hence, by AA, . So Hence . If we add to both sides we see that .

Theorem. In the three medians , , and meet in a point. Proof. Let G be the point of intersection of and . We need to show that G also lies on . In order to do this we let the intersection point of and , and we will show that .

By the lemma. However, we can also apply the lemma to By the lemma . However, we can also apply the lemma to . is the intersection point of two medians in a triangle and so it must also trisect them. Hence, But only has one trisection point closer to so Corollary. The center of gravity trisects each of the medians.

Length Formulas In this section we will calculate the lengths of the medians of . We will denote these lengths by , and . Theorem. Proof. Let be the altitude. We assume for convenience that D is between B and , as shown in the Figure. Now apply the geometric law of cosines to the triangles, and to obtain

and or Adding, we get as claimed.

Of course, there are similar formulas for the other two medians: and With a bit of algebra we can deduce a number of striking formulas from these. Corollary. (a) (b) (c)

Complementary and Anticomplementary Triangles Let be a triangle and let the midpoints of the sides be , , and . Then the triangle with these three points for vertices is called the complementary or median triangle of .

Theorem. Given with complementary triangle , (1) The sides of are parallel to the sides of . (2) and are similar with ratio . (3) and have the same centroid G. Proof. (1) We showed before that and are parallel. The same proof shows that and are parallel and and are parallel. (2) we also proved that . The same proof shows that and . Hence by SSS for similar triangles.

(3) Let intersect at the point M. Consider the triangle and (3) Let intersect at the point M. Consider the triangle and . Since , by the corresponding angles theorem. Likewise, . So the two triangles must be similar by AA. Hence Which implies that . But so . Putting these together we see that and that M is the midpoint of . With this observation in hand the rest of the proof follows easily. is a median of , is a median of . and .

Similar statements are true for the other two medians Similar statements are true for the other two medians. Hence, the point of intersection of the three medians of is the same as the point of intersection of the three medians of . Next, given a triangle , we define the anticomplementary triangle of to be the unique triangle with the properties that contains the point A and is parallel to , contains the point B and is parallel to , and contains the point C and is parallel to .

Theorem. If a triangle has anticomplementary triangle , then is the complementary triangle of . Proof. To prove this theorem it is worthwhile to recall the definition of a complementary triangle. In order to prove that is the complementary triangle of we need to show (see the Figure) that A is the midpoint of and C is the midpoint of . We will only prove that A is the midpoint of since the other two cases are similar. Because and , the quadrilateral must be a parallelogram. We know that opposite sides of a parallelogram are congruent, so .

Likewise, since and , the quadrilateral Likewise, since and , the quadrilateral . is also a parallelogram and consequently Comparing these two congruences, we see that , which means that A must be the midpoint of . This is what we wanted to show. Corollary. 1. and are similar with ratio 2. 2. and have the same centroid.

The Orthocenter Previously, we proved that in any triangle the three medians meet in a point G, the three angle bisectors meet in a point I, and the three perpendicular bisectors of the sides meet in a point O. Here we will prove the corresponding theorem for the altitudes, namely that the three altitudes meet in a point. The point is called the orthocenter of , and we will denote it by H. We point out that H does not necessarily lie inside the triangle: If is a right triangle, then H is at the vertex, and if is obtuse, then H will lie outside.

Theorem. If a triangle has altitudes , , and , then , , and meet at a point. Proof. Let be the anticomplementary triangle of . Since A is the midpoint of and since is perpendicular to , we conclude that is the perpendicular bisector of .

Similarly, is the perpendicular bisector of and is the perpendicular bisector of . But we know that the three perpendicular bisectors of the sides of any triangle meet in a point. Hence, , , and meet in a point, as claimed.

In addition to being short, this proof yields two corollaries In addition to being short, this proof yields two corollaries. Their proofs follow from the fact that the perpendicular bisectors of the sides of a triangle meet at the circumcenter of that triangle. Corollary. Let be a triangle with anticomplementary triangle . Then the orthocenter of is the circumcenter of . Corollary. Let be a triangle with complementary triangle . Then the circumcenter of is the orthocenter of .

Fagnano’s Problem Here we will solve a geometric optimization problem due to G. C. Fagnano in 1775; Fagnano also solved it. But the proof we shall give is due to L. Fejer in 1990. Problem. Given an acute triangle , find points X on , Y on , and Z on that minimize the perimeter of the resulting triangle .

We claim that is the required triangle. First Step: Given an acute angled triangle and given a fixed point X on , then find points Y on and Z on that minimize the perimeter of .

A B C 2 1 Z X Y 4 3 F E SAS SAS SAS SAS

Second Step: We have shown that how to choose Y and Z if we are given X. Our next goal is to determine a choice of X. As before, let X be on , be the reflection of X through , and the reflection of X through . We have shown that the minimum perimeter of a triangle inscribed in with X as a vertex is the length of . What we have to do now is find that particular X that minimizes the length . Since A is on the perpendicular bisector of , and since A is on the perpendicular bisector of , , for any choice of X on .

Also, using the fact that is isosceles and has congruent base angles, ; and since is isosceles, we have Adding, we see that

Now consider . This triangle has three properties: 1. , a relation that does not depend upon the choice of X. 2. The sides and are congruent to each other and to . 3. The length of the base is the perimeter of , which we want to minimize. Since is an isosceles triangle with a fixed summit angle, we minimize the base by minimizing the length of the legs . But this is the length AX.

Now we are as good as done Now we are as good as done. We know that of all the line segments joining A to a point on the shortest one is the perpendicular, . Therefore X = D, will be a vertex of the triangle that solves Fagnano’s problem. By similar arguments the other two vertices will be E and F and the orthic triangle is the one of smallest perimeter. We remark that has perimeter ,where K = area and R = circumradius.

The Euler Line Recall the notation O = the circumcenter, G = the centroid, and H = the orthocenter of . We now have the follwing remarkable theorem, due to Euler. Theorem. The three points O, G, and H lie on a straight line, with G between O and H. Moreover, GH = 2GO.

Proof. As usual, we let be the midpoint of and the altitude to as in the Figure. Then is a median, it contains G, and is the perpendicular bisector of Connect O to G and let the resulting line intersect at the point P. Since and are each perpendicular to , they must be parallel.

Hence, by the alternate interior angle theorem and Hence, by the alternate interior angle theorem and . By AA, we conclude that So But the centroid G trisects the median , so Hence This equation is the key to the proof. Consider what we just proved from the point of view of O and G. We have shown that if you extend the line segment to a new point P such that G is between O and P and , then this point P must be on the altitude , since these two conditions determine P uniquely.

However, the same manner of proof would show that this point P will be on the altitude and the altitude . Thus, P is the orthocenter H and the theorem is proved. The line joining O, G, and H is called the Euler Line of