Homework 8 solutions

Notes:

These solutions are presented without diagrams. To understand the solutions you should draw and label your own diagrams! In problem 1 careful distinctions are made between angle measure, proper angle measure, and the absolute value of proper angle measure. To get the idea of the proof you should ignore these distinctions (as we did in class discussion). In other words, pretend that y1, y1', and |y1'| are all the same thing. No solution is given here for the extra credit problem 3.

1. If two opposite angles of a quadrilateral are supplementary, then the quadrilateral can be inscribed in a circle.

Proposition: The sum of the angles of a quadrilateral is 360 degrees.

Proof. Let ABCD be a quadrilateral, and let a = a(DAB), b = a(ABC), c = a(BCD), d = a(CDA). We will prove that a + b + c + d = 360. The diagonal BD divides
the quadrilateral into two triangles ABD and BCD. Let s = a(ABD), t = a(DBC), u = a(BDA), v = a(CDB). By the angle addition axiom, b = s + t and d = u + v. By
the triangle angle sum theorem, a + s + u = 180 and c + t + v = 180. Therefore,

(a + s + u) + (c + t + v) = 360

a + (s + t) + c + (u + v) = 360

a + b + c + d = 360

QED

Now we can prove the given theorem.

First we prove that if a quadrilateral can be inscribed in a circle, then its opposite angles are supplementary.

Suppose the quadrilateral ABCD is inscribed in the circle X with center O. As above, let a, b, c, d be the angles of ABCD. Consider the radii OA, OB, OC, OD of
X. Since these four segments have equal lenght (the radius of the circle), the four triangles OAB, OBC, OCD, ODA are isosceles. By the isosceles triangle theorem,
the two base angles of each of these triangles are equal. Let w = a(OAB) = a(ABO), x = a(OBC) = a(BCO), y = a(OCD) = a(CDO), z = a(ODA) = a(DAO). By the
angle addition axiom, a = z + w, b = w + x, c = x + y, d = y + z. By the above proposition,

a + b + c + d = 360

(z + w) + (w + x) + (x + y) + (y + z) = 360

2w + 2x + 2y + 2z = 360

w + x + y + z = 180

(w + x) + (y + z) = 180

b + d = 180

Thus the opposite angles b and d are supplementary.

Also,

(z + w) + (x + y) = 180

a + c = 180.

Thus the opposite angles a and c are also supplementary.

Next we prove that if the opposite angles of a quadrilateral are supplementary, then the quadrilateral can be inscribed in a circle.

Suppose that the quadrilateral ABCD, with angles a, b, c, d as above, has a + c = 180 and b + d = 180. Let X be the circumcircle of the triangle ABC, and let O be
the center of X. (Recall that we have proved that the circumcircle of a triangle exists - its center is the intersection of the perpendicular bisectors of the sides of the
triangle.) Since A, B, C lie on X, the lengths of the segments OA, OB, OC are equal to r, the radius of X. Consider the segment OD. We will prove that D lies on X
by showing that the length of OD is also equal to r. Let s = OD.

The two triangles OAB and OBC are isosceles, and so the two base angles of each of these triangles are equal. Let w = a(OAB) = a(ABO), x = a(OBC) = a(BCO), y1
= a(OCD), y2 = a(CDO), z1 = a(DAO), z2 = a(ODA). Since a + c = 180, we have (z1 + w) + (x + y1) = 180. Since b + d = 180, we have (w + x) + (y2 + z2) = 180.
Therefore y1 + z1 = y2 + z2.

Now let y1' = a'(OCD), y2' = a'(CDO), z1' = a'(DAO), z2' = a'(ADO). These four numbers have the same sign and |y2'| + |z2'| < 180. Thus |y1'| + |z1'| = |y2'| + |z2'|.

By the angle inequality, if s < r then |y2'| > |y1'| and |z2'| > |z1'|. Thus |y2'| + |z2'| > |y1'| + |z1'|, which is a contradiction.

On the other hand, the angle inequality implies that if s > r then |y2'| < |y1'| and |z2'| < |z1'|. Thus |y2'| + |z2'| < |y1'| + |z1'|., which is a contradiction.

Therefore s = r, so D lies on the circle X, which is what we wanted to prove.

2. The medians of a triangle are concurrent.

We will prove that if ABC is a triangle and AD is a median, then both of the other two medians intersects AD at the point O of AD such that AO = 2 DO. Therefore
the three medians are concurrent at the point O.

Consider the median BE. (The same proof will apply to the median CF.) The betweennes axiom implies that ray AD is between ray AB and ray AC. Since ray AC =
ray AE, the betweenness axiom implies that ray AD intersects line BE at a point between B and E. The same argument shows that ray BE intersects line AD at at
point between A and D. Thus the segments AD and BE intersect at a point O.

Consider the line DE. Since D is the midpoint of BC and E is the midpoint of AC, the side-angle-side similarity theorem implies that the triangles ECD and ACB are
similar, with similarity ratio 2. Thus 2 ED = AB.

It also follows from the similarity of these two triangles that a(CED) = a(CAB), which implies that the lines ED and AB are parallel (by the corresponding angles
theorem). Thus a(DEO) = a(ABO) and a(ODE) = a(OAB) (by the alternate interior angles theorem). Therefore triangles OED and OBA are similar, with similarity
ratio 2, by the angle-side-angle similarity theorem. Thus AO = 2 DO, which is what we wanted to prove.