Math 444 Lab 6, 11/7/2001

Start by carrying out the constructions of 4, 6, 7, 8.

Concurrence of lines in triangles

This overview of concurrence results is partly review and partly new.  You can skip the review figures if you wish.

1. Concurrence of medians (review)

In a triangle ABC, let the midpoints of the sides be A', B' and C' so that the medians are AA', BB', CC'.  Then the medians are concurrent at a point G called the centroid.  The point G has the property that AG/AA' = BG/BB' = CG/CC' = 2/3.

The proof starts by intersecting two medians AA' and BB' at a point P and showing that AP/AA' = BP/BB' = 2/3. The tools for this are parallel lines and ratios.  See hint in B&B problem 44, page 259 or an earlier lab.

Then intersect the two medians BB' and CC' at a point Q with CQ/CC' = BQ/BB' = 2/3.  The fact that BQ/BB' = BP/BB' implies P = Q.

2. Ceva's Theorem (review)

This is a more general result than the concurrence of medians.  Let ABC be a triangle and let A', B' and C' be points on sides BC, CA, and AB respectively.  Then the lines AA', BB' and CC' are concurrent at a point P if and only if

(BA'/A'C)(CB'/B'A)(AC'/C'B) = 1

A couple of proofs were given for this.  One approach was to look for masses at the vertices for A, B, and C so that P is the center of mass.  In this case, each of the ratios such as BA'/A'C is a ratio of two masses.  A second approach uses ratios of areas instead of ratios of masses.

Note:  A line through a vertex of a triangle is called a cevian line; thus ceva's theorem is about concurrence of cevian lines.  The concurrence of medians is a corollary of Ceva's Theorem.  Ceva's theorem and the concurrence of medians are examples of theorems in affine geometry, since the ratio relationships stay the same if the figure is projected to another plane by parallel projection.

3. Concurrence of perpendicular bisectors of a triangle (review)

Given a triangle ABC, the perpendicular bisectors of the sides are concurrent at a point O.  The point O is called the circumcenter of the triangle and O is the unique point equidistant from all 3 vertices, so the circle through A, B, C is centered at O.  This circle is called the circumcircle.

The proof follows from intersecting two perpendicular bisectors at a point P and observing that P is equidistant from all 3 vertices, so it is on all 3 bisectors by the Locus Property of Perpendicular Bisectors (Principle 10 of B&B).

4. Concurrence of Altitudes of a triangle (NEW!)

Given a triangle ABC, construct the 3 altitudes of the triangle and observe that they APPEAR to be concurrent.  How can we explain this?

One quick way to see this is to exploit a connection between altitudes and perpendicular bisectors.  Here is how it goes.

Construct the triangle ABC and its altitudes.  Then through each vertex, construct the line that is parallel to the opposite side of the triangle.  These 3 new lines form a new, bigger triangle A1 B1 C1 (in this case A1 is the intersection of the two parallels through B and C, etc.).

• What is the relationship between ABC and the new triangle A1B1C1?
• The altitudes in ABC are also special lines for A1B1C1.  What are they?
• How can we use this figure and what we know already to conclude that the altitudes of ABC are concurrent?

The point of concurrence of the 3 altitudes of ABC is called the orthocenter of ABC.

SAVE THIS FIGURE.

5. Concurrence of the (interior) Angle Bisectors of a triangle (Review)

Given a triangle ABC, construct the bisectors of each angle.  Then these bisectors are concurrent at a point I inside the triangle.  The point I is called the incenter of the triangle.  It is the center of the unique circle inscribed inside the triangle.

The proof uses the idea that if a point is on the angle bisector, then the point is equidistant from the two sides of the angle.  Then the proof follows the logic of the perpendicular bisector proof.

6. Concurrence relations of interior-exterior Angle Bisectors of a triangle (sort of new)

• Given a triangle ABC, extend the sides to LINES, not segments. 
• Construct the bisectors of each angle.  Color the bisectors blue.
• Construct at each vertex the perpendicular line to the angle bisector at that vertex.  Color these lines red. Explain why each such line bisects the exterior angles at a vertex.
• Now observe that in addition to the point of concurrency I of the interior angle bisectors, any two exterior bisectors intersect at a point on the interior angle bisector of a vertex.  How many points of concurrency of angle bisectors are there altogether?
• The points of concurrency outside the triangle ABC are called the excenters of the triangle.  Label the excenters XYZ.  Each excenter is the center of a circle tangent to all three LINES AB, BC, CA.  Explain why.

7. Connection between Altitudes and Angle Bisectors (NEW)

Continue with the angle bisector figure above.

• Notice that the interior angle bisectors of ABC are rays on special lines of triangle XYZ.  What are these lines?  How does this give a connection between two concurrency facts for triangles?

Now return to the altitude figure from #4. 

• Construct the feet of the altitudes.  Call the feet PQR.  The triangle PQR is called the orthic triangle of ABC. 
• Notice that the altitudes are the interior angle bisectors of PQR and the sides of ABC are the exterior angle bisectors.

[Note: we have not quite proved this.  We have found the relation starting with PQR but not starting from ABC.]

8. Isosceles Trapezoids (sort of NEW)

Construct a figure ABCD so that AB = CD and angle ABC = angle BCD (and A and D are on the same side of line BC).  Make this dynamic so that you can drag A, B and/or C.

The figure ABCD is called an isosceles trapezoid.  Explain why line AD is parallel to BC.  Explain why the perpendicular bisector of BC is a line of symmetry of this figure.

Now construct a circle through A, B, C, D.  Also construct a circle tangent to line AB, segment BC, and line CD.  What kind of concurrence occurs here?