1. Opposite angles of a quadrilateral are supplementary if and only if the quadrilateral can be inscribed in a circle. (Half of this theorem was used in exam 2.)

If opposite angles of a quadrilateral are supplementary then the other two opposite angle are supplementary.

In figure ABCD (quadrilateral), angle ABC is opposite to ADC and angle BAD is opposite to DCB if these opposite angle are supplementary, then figure 1 can be inscribed in a circle. We can carry out this investigation by constructing a diagonal AC and create triangles ABC and ACD as shown in figure 1 above. By using the figure above, let the measure of these angles be as follows: a(ABC) = x, a(CAB) = y, a(ACB) = z, a(CAD) = v, a(ACD) = p, a(ADC) = h. By using the theorem of the triangle sum, x + y + z = 180 degrees and v + h + p = 180 degrees. By using the angle axiom, we can show all the measure of all interior angles of a quadrilateral. Note that a(BAC) = y + v, a(BCD) = z + p, therefore the measure of all interior angles of the quadrilateral are x + (y + v) + h + (z + p) = 180 + 180 = 360 degrees total. Therefore, If angles BAD and BCD are supplementary, then their sum is 180 degrees, which means angle x and y = 360 - 180 = 180 degrees which are supplementary.

Proof:

As noted from above that if two opposite angles are supplementary, then the other pair of opposite angles must also be supplementary. Therefore from figure 2, a(ABC) + a(ADC) = 180, a(BAD) + a(DCB) = 180 degrees. Therefore no matter what pair of opposite angles I choose, they must be supplementary.

Now, construct segment BD and create triangles BDC and BAD as shown in figure 3. Since I define a quadrilateral to be formed by four vertices that are not collinear, and then choose any three of them and from homework #7 we noticed that their exist a unique circle passing through the three points. Now with the three points BDA, construct a circle passing through the three points and o being the center of this circle with oD, oB and oA as the radii of the circle. By definition of a circle, oD, oB and oA are equal hence triangles AoD, BoA and DoB are all isoceles triangles. By isoceles triangle theorem, a(oAB) = a(oBA) = x, a(oAD) = a(oDA) = y, a(oDB) = a(oBD) = z. By angle axiom, and triangle angle sum theorem, the sum of the angles of triangle BAD are 2x + 2y + 2z, which must = 180 degrees. By same reasoning, a(BoA) = 2y + 2z, a(AoD) = 2x + 2z and a(DoB) = 2x + 2y. By the angle axiom, a(ABD) + a(ADB) = x + y + 2z. Therefore, by the given and the angle axiom, a(CBD) + a(CDB) must equal x + y. Let a(CBD) = a and a(CDB) = b. So by substitution, a + b = x + y

By using the same figure (figure 4), choose non collinear points A,B, and C. Again, by the theorem from homework # 7, unique circle passes through these points. Construct it. Let o' be the center of this circle. Construct these radii of the circle: segments o'B, o'C and o'D. From above, a(CBD) = a and a(CDB) = y. By definition, angles CDB and CBD are inscribed in the circle, and angles Bo'C and Co'D are central angles. Angles CBD and Co'D intercept the same arc, and angles CDB and Bo'C intercept the same arc. Therefore, we can state that a(Bo'C) = 2b and a(Co'D) = 2a. By angle axiom and then by substitution, a(Bo'D) = 2a + 2b = (a+b) + (a+b) = (x+y) + (x+y) = 2x + 2y. By the triangle angle sum theorem, the sum of the angles in triangle Bo'D must equal 180 degrees, or 2x + 2y + 2z. Therefore, by the angle axiom and the triangle angle sum theorem, a(o'BD) + a(o'DB) = 2z. But, since d(o'B) = d(o'C) = d(o'D) by definition of a circle, then triangle Bo'D must be isoceles. So that means a(o'BD) = a(o'DB) = z.

Finally from figure 3 and 4 we can bring everything together. a(DBo) = a(BDo) = a(o'BD) = a(o'DB) = c, and side BD is reflexive. Therefore, triangle BoD and Bo'D are congruent by ASA theorem. That means that d(oB) = d(oD) = d(o'B) = d(o'D) by transitivity. Therefore, o and o' are the same point. Thus, d(oB) = d(oC) = d(oD) = d(oA). Since all of the vertices of the quadrilateral are equidistant from the same point o, then o serves as the center of a circle which passes through all of the vertices. This fact, by definition, means that the quadrilateral can be inscribed in a circle, which was to proved. For more demonstration, click here.

2. The medians of a triangle are concurrent. (The idea of this proof has been discussed in class.)

To proof that medians are concurrent, I will use figure 1 and the properties of medians. Medians are line segments connecting the midpoint of a side of a triangle and the vertex opposite to that side as shown in figure 1.

• Median intersect a triangle into two triangle whose areas are congruent.
• The ratio of intersection of the medians is 2 : 1.

In figure 1 construct segment ED by midpoint theorem, it can be noted that segment ED is parallel to AB. Now consider triangle ABC and EDC, it should be noted that both triangles share angle BCA. Angles ABC and EDC are corresponding hence equal. Finally angles BAC and angle DEC are corresponding, hence equal. Now, by angle angle angle similarity, triangle ABC is similar to EDC. which implies that

Now consider the two triangles (ABH and HED), a(DEB) = a(EBA), and a(BAD) = a(ADE) since they are corresponding angles. Also a(EHD) is equal to a(AHB) by triangle angle sum. Hence by angle angle angle similarity, we can state that triangle DEH is similar to ABH. Finally, since DE/AB = 1/2 from eq 1, and triangle ABH is similar to EHD, then which means that median AD and BE divide each other in the ratio of 2 : 1 which will also apply to other median of the triangle. Hence medians are concurrent. click here to see the gsp file.

Well, the investigation looks tricky! but let me try. It should be noted that their are two cases to consider,

• when the lines are concurrent
• when the lines are parallel

when the lines are concurrent

By aid of instructions given, I will construct the points such that A, B and C are noncollinear and join them to form triangle ABC.Construct another point not equal to any of the three and name it O. Join the three points of the triangle by a line to point O as shown in figure 1. Next Pick a point on line AO and name it A' then construct two lines passing through A' and one parallel to line AB (to meet BO at B') then the other parallel to AC (to meet CO at C'). Then from C' draw a line parallel to BC and through B' to form triangle A'B'C' ok!

In my strategy, I will like to proof:

• Triangle ABC is similar to A'B'C'
• lines from ABC to O, and lines from A'B'C' to O are the same lines if so, then I will conclude that they are concurrent.

Take a look at triangle ABC and A'B'C', line BC is parallel to B'C' which means angle ABC is corresponding to angle A'B'C'. Also, line AC is parallel to line A'C' which means that angle BAC is corresponding to angle B'A'C', hence equal. By the angle sum theorem, angle BCA is qual to angle B'C'A'. Therefore by angle angle angle similarity, triangle ABC is similar to

A'B'C' hence: AB/A'B' = BC/B'C' = AC/A'B' = K .. (eq. 1)

Now consider triangle BOC and B'OC' in figure 2. It should be noted that both triangle share same angle i.e angle BOC and angle B'OC'. Therefore angle BOC is equal to angle B'OC'. using CO as the transversal line and given line CB is parallel to line C'B', the angle B'C'O is corresponding angle to BCO hence equal. Using angle sum of a triangle, angle OB'C' is equal to OBC. By angle angle angle similarity theorem, triangle BOC is similar to triangle B'OC'. Now then, BC/B'C' = BO/B'O = CO/C'O = k ... (eq. 2), therefore line BO & B'O, CO & C'O, and AO & A'O are the same lines. Hence they meet at the same point O. Click here to see gsp file

When the lines are parallel

Given AB // A'B', AC // A'C' and BC // B'C', then the guess is the figure BB'C'C is a parallelogram. If that is the case, then the sides of the figure are parallel. To show this, I will use the parallel axiom to show that line BB' is parallel to CC' and line BC is parallel to B'C'. Consider angle gB'B and gC'C, they are corresponding hence equal. If this angles are equal, then angle BB'C' is supplimentary to angle B'C'C, where B'C' is the transversal line. Which means a(BCC') + a(B'C'C) = 180 degrees. Therefore by using parallel axiom, line CC' is parallel to BB'.

With the same argument, AA' is also parallel to BB" and CC' click here for gsp file.

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