Definition 3.3.1. Median.
A line is a median if and only if it connects a vertex of a triangle to the midpoint of the opposing side.
Section 3.3 Concurrent
Subsection 3.3.1 Explore
Geogebra will be helpful for performing these experiments. Be as detailed as you can with your conjectures.
Checkpoint 3.3.2.
Use the Geogebra example in Figure 3.3.3 to experiment with the relationship of the three perpendicular bisectors of a triangle. Move the vertices of the triangle around. What remains true about the perpendicular bisectors?
Instructions.
Drag the vertices of the triangles around and see what is always true about the perpendicular bisectors.
Checkpoint 3.3.4.
Use the Geogebra example in Figure 3.3.5 to experiment with the relationship of the three medians of a triangle. Move the vertices of the triangle around. What remains true about the medians?
Solution.
- The medians intersect at one (1) singluar point regardless of how the triangle is manipulated.
- The medians will always form from the midpoint of each side to the common intersection point.
Instructions.
Drag the vertices of the triangles around and see what is always true about the medians.
Checkpoint 3.3.6.
Use the Geogebra example in Figure 3.3.7 to experiment with the relationship of the three angle bisectors of a triangle. Move the vertices of the triangle around. What remains true about the angle bisectors?
Solution.
- Angle bisectors always divide the triangle into 6 smaller triangles. This is unlike the perpendicular bisectors that can divide the triangle into fewer areas that don't need to be triangular.
- The intersection of the angle bisectors is always IN the triangle.
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When the three points of the triangle \(A, B, C\) are colinear so \(A - B - C\text{,}\) the angle bisectors form perpendicular lines that intersect at \(B.\)
- It appears that this is the only case in which two angle bisectors can form a right angle.
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The areas of the 6 triangles formed by the angle bisectors do not all need to be equal. Further it is difficult to make the areas equal, but an equilateral triangle has all triangle areas equal.
Instructions.
Drag the vertices of the triangles around and see what is always true about the angle bisectors.
Checkpoint 3.3.8.
Use the Geogebra example in Figure 3.3.9 to experiment with the relationship of the three altitudes of a triangle. Move the vertices of the triangle around. What remains true about the altitudes?
Solution.
The altitudes always intersect at a single point, but that point may lie in the interior, boundary, or exterior of the triangle. (In the degenerate case, the altitudes are parallel, as they were with the perpendicular bisectors.)
Instructions.
Drag the vertices of the triangles around and see what is always true about the altitudes.
Checkpoint 3.3.10.
Construct \(\triangle ABC.\) Construct \(\triangle XYZ\) such that \(X-B-Y,\) \(Y-C-Z,\) \(Z-A-X,\) and \(\overline{XY} \parallel \overline{AC},\) \(\overline{YZ} \parallel \overline{AB},\) \(\overline{ZX} \parallel \overline{BC}.\) Construct the perpendicular bisectors of \(\triangle XYZ.\) What appears to be true of these with respect to \(\triangle ABC.\)
Solution.
When we perform the Checkpoint's steps, we have the given visual as an example. Firstly, all perpendicular bisectors intersect at some point, either on the interior or exterior of the triangle \(\triangle ABC\text{.}\) At first glance, depending on how you construct the triangle, it may seem that some of the perpendicular bisectors of \(\triangle XYZ\) are parallel to some of the sides of \(\triangle ABC\text{,}\) but zooming out will eventually disprove this claim. However, it seems more likely that \(\triangle XYZ\)'s perpendicular bisectors intersect the sides of \(\triangle ABC\) at a \(90^\circ\) angle. Finally, if we shift one of the triangle's points across another point, one of the sides on \(\triangle XYZ\) and its corresponding perpendicular bisector make a complete rotation by the time the shifting is complete (in the case that the points lie on top of one another, two of the sides of \(\triangle XYZ\) lie on top of one another, and two of the corresponding perpendicular bisectors also lie on top of one another.
Subsection 3.3.2 Prove
Lemma 3.3.11.
Consider three points \(A,B,C\) with \(\ell_1\) and \(\ell_2\) the perpendicular bisectors of \(\overline{AB}\) and \(\overline{BC}\) respectively. Let \(M_2=\ell_2 \cap \overline{BC}.\) Show \(\ell_1 \parallel \ell_2\) implies the existence of \(D=\ell_2 \cap \stackrel{\longleftrightarrow}{AB}\) such that \(A,\) \(B,\) and \(D\) are collinear and \(\angle BDM_2\) is a right angle.
Proof.
Consider three points \(A, B\text{,}\) and \(C\) such that \(\ell_1\) and \(\ell_2\) are the perpendicular bisectors of \(\overline{AB}\) and \(\overline{BC}\) respectively. Let \(M_2 = \ell_2 \cap \overline{BC}\) be the midpoint of \(\overline{BC}\text{.}\) Finally, let \(\ell_1\) and \(\ell_2\) be parallel. Note because \(\ell_1 \parallel \ell_2\) and since they respectively perpendicularly bisect \(\overline{AB}\) and \(\overline{BC}\text{,}\) then \(\overline{AB}\) has to eventually intersect \(\ell_2\) using Ruler and Distance Postulates. This is because considering the point \(M_1 = \ell_1 \cap \overline{AB}\text{,}\) Playfair suggests there is only one line passing through \(M_1\) parallel to \(\ell_2\text{,}\) so \(\overleftrightarrow{AB}\) and \(\ell_2\) will intersect. Call this intersection point \(D = \overleftrightarrow{AB} \cap \ell_2\text{.}\) Because we've only extended \(\overleftrightarrow{AB}\text{,}\) we know \(A, B\text{,}\) and \(D\) are collinear. As well, note the line \(\ell_2\) contains both \(M_2\) and \(D\text{.}\) Because \(\ell_1 \parallel \ell_2\) imply the right angle formed by \(\ell_1\) and \(\overline{AB}\) equals \(\angle BDM_2\) by Alternate Interior Angles, then \(\angle BDM_2\) is a right angle. So, we have shown given \(\ell_1 \parallel \ell_2\text{,}\) that there exists \(D = \overleftrightarrow{AB} \cap \ell_2\) such that \(A, B\text{,}\) and \(D\) are collinear and \(\angle BDM_2\) is a right angle. Note the lemma is proven, but such construction is impossible under Euclidean geometry.
Theorem 3.3.12. Perpendicular bisectors.
Prove the conjecture about the perpendicular bisectors.
Proof.
Consider non-collinear points \(A, B\text{,}\) and \(C\text{,}\) such that the line segments \(\overline{AB}, \overline{BC}\text{,}\) and \(\overline{AC}\) form a triangle \(\triangle ABC\text{.}\) First, make the perpendicular bisectors of two sides, \(\overline{AB}\) and \(\overline{BC}\)—respectively called \(\ell\) and \(m\text{.}\) Note because \(A, B\text{,}\) and \(C\) are non-collinear, \(\ell\) and \(m\) are not parallel so they intersect at some point \(X\text{.}\) Note \(X\) does not necessarily need to be on the triangle's interior. Anyways, Theorem 2.1.9 states a point is on the perpendicular bisector of a segment if and only if it is equidistant to the endpoints. Thus, because point \(X\) lies on the perpendicular bisector of \(\overline{AB}\) and \(\overline{BC}\text{,}\) then \(\overline{XA} = \overline{XB}\) and \(\overline{XB} = \overline{XC}\text{.}\) This implies \(\overline{XA} = \overline{XC}\text{,}\) and so we can use the biconditional of Theorem 2.1.9 to say \(X\) is on the perpendicular bisector of \(\overline{AC}\text{.}\) Therefore, all three perpendicular bisectors intersect at a point.
Theorem 3.3.13. Circumcenter.
Three points uniquely determine a circle.
Proof.
Consider triangle \(\triangle ABC\text{.}\) Let the midpoints of the sides be \(D\text{,}\) \(E\text{,}\) and \(F\text{,}\) respectively. Then the perpendicular bisectors of the three sides pass through \(D\text{,}\) \(E\text{,}\) and \(F\text{,}\) respectively.
Next consider any two perpendicular bisectors, say the bisectors of sides \(\overline{AB}\) and \(\overline{AC}\text{.}\) These bisectors intersect each other somewhere by repeated application of Pasch's axiom. Call their intersection \(G\text{.}\) Since \(G\) lies on the perpendicular bisector of each segment, it is equidistant from \(A\) and \(B\text{,}\) as well as from \(A\) and \(C\text{.}\) That means that the lengths \(AG\text{,}\) \(BG\text{,}\) and \(CG\) are all equal. Thus \(G\) is also equidistant from both \(B\) and \(C\text{,}\) so \(G\) lies on the perpendicular bisector of \(\overline{BC}\text{.}\)
Therefore, the three perpendicular bisectors of the sides of \(\triangle ABC\) meet at a single point.
Lemma 3.3.14.
Two medians intersect at a point \(2/3\) of the way down both medians.
Proof.
Given \(\Delta ABC\text{,}\) we can construct two medians, BE and CD. DE is parallel to BC and DE is half the length of BC since D and E are midpoints of AB and AC, respectively. \(Delta DEF\) is similar to \(\Delta BCF\text{.}\)
\begin{align*}
\frac{EF}{BF} = & \frac{DE}{BC} = & 1/2. \\
EF = & 1/2(BF) \\
BE = & 1(BF) + 1/2(BF) = & 3/2(BF) \\
BF = & 2/3(BE)
\end{align*}
Similarly,
\begin{align*}
\frac{DF}{CF} = & \frac{DE}{BC} = & 1/2 \\
DF = & 1/2(CF) \\
CD = & 1(CF) + 1/2(CF) = & 3/2(CF) \\
CF = & 2/3(CD)
\end{align*}
Theorem 3.3.15. Medians.
Prove the conjecture about the medians.
Proof.
Construct \(\triangle ABC\text{.}\) Let the midpoints of the sides of this triangle be \(D\text{,}\) \(E\text{,}\) and \(F\text{.}\) Extend the medians \(\overline{AE}\text{,}\) \(\overline{BF}\) and \(\overline{CD}\text{.}\)
Consider any two of the medians of \(\triangle ABC\text{,}\) say \(\overline{BF}\) and \(\overline{CD}\text{.}\) By repeated application of the Crossbar theorem, these medians intersect. Let their intersection be \(G\text{.}\) Next construct the line through points \(D\) and \(F\text{.}\) This line divides two sides of \(\triangle ABC\) proportionally, because \(D\) and \(F\) are the midpoints of sides \(\overline{AB}\) and \(overline{AC}\) respectively. By a previous theorem, this line is parallel to side \(\overline{BC}\text{.}\)
Next examine the triangles \(\triangle GBC\) and \(\triangle GFD\text{.}\) (The line \(DF\) contains a side of the second triangle.) We claim these triangles are similar. The angles at \(G\) are congruent because they are vertical angles. Then \(\angle FDG \cong \angle BCG\) by alternate interior angles, and likewise for \(\angle DFG\) and \(\angle GBC\text{.}\)
Since line \(DF\) divides the sides of \(\triangle ABC\) in half, the segment \(\overline{DF}\) has half the length of segment \(\overline{BC}\text{,}\) by triangle similarity. Thus the sides of \(\triangle GFD\) have half the length of the sides of triangle \(GCB\text{.}\) It follows that \(|CD| = 3|DG|\text{,}\) so \(|DG| = (1/3)|CD|\text{.}\) Since \(|CD| = |DG| + |GC|\text{,}\) we must have \(|CD| = (1/3)|CD| + |GC|\text{,}\) which implies that \(|GC| = (2/3)|CD|\text{.}\) So \(G\) lies two-thirds of the length down the median \(CD\text{.}\) Identical reasoning applies to the other medians.
Theorem 3.3.16.
A point is on the angle bisector of an angle if and only if it is equidistant from both sides of the angle.
Proof.
First, we will prove that the point P on the angle bisector implies equidistance. \(\overrightarrow{AP}\) bisects \(\angle BAC \text{.}\) Since the distance between a point and a line is from the point to the foot of the perpendicular, \(m\angle PBA \cong m\angle PCA = \frac{\pi}{2}. \) By AAS, \(\Delta BAP \cong \Delta CAP \) By CPCTC, \(BP \cong CP\) and thus the point P is equidistant from AB and AC.
Second, we will prove that equidistance implies that the point P is on the angle bisector. P is a point in the interior of the angle BAC such that \(|BP| = |CP|\text{.}\)
\(\angle PBA \cong \angle PCA = \frac{\pi}{2} \) By RSS, \(\Delta BAP \cong \Delta CAP \) By CPCTC, \(\angle BAP \cong \angle CAP \) so P is on the angle bisector of \(\angle BAC\text{.}\)
Theorem 3.3.17. Angle bisectors.
Prove the conjecture about the angle bisectors.
Proof.
First note that any two angle bisectors of \(\triangle ABC\) must intersect somewhere within the triangle by Crossbar: Since an angle bisector intersects a vertex of the triangle and there exists some point on the bisector within the triangle, the bisector must intersect the opposite side of the triangle at a point \(X\text{.}\) Apply the same reasoning to the triangle \(\triangle ABX\) to see that the other bisector must intersect the first.
We now show that each pair of bisectors intersects at the same point. Consider \(\triangle ABC\) and let \(G\) be the intersection of the angle bisectors of \(\angle ABC\) and \(\angle ACB\text{.}\) By the preceding theorem, \(G\) is equidistant from lines \(AB\) and \(BC\) and also from lines \(BC\) and \(AC.\) Note then that \(G\) must be same distance from all three lines. Let \(D\text{,}\) \(E\text{,}\) and \(F\) be the points at the foot of the perpendicular from \(G\) to segments \(\overline{AB}\text{,}\) \(\overline{BC}\text{,}\) and \(\overline{AC}\) respectively. Then we have shown that \(|DG| = |EG| = |FG|\text{.}\)
But then \(G\) is equidistant from the two sides of \(\angle BAC\text{,}\) and so it lies on the angle bisector of that angle as well. Therefore, the three angle bisectors of a triangle intersect at a single point.
Theorem 3.3.18. Incenter.
For each triangle there exists a circle inside and tangent to all three sides.
Theorem 3.3.19. Altitudes.
Prove the conjecture about the altitudes.
Proof.
We begin with triangle ABC. The altitudes of the triangle are AD, BE, and CF. We will show that the altitudes are concurrent. We construct triangle GHI such that H-A-G, H-B-I, G-C-I, and GH is parallel to BC, HI is parallel to AC, and GI is parallel to AB.
BC = AH because AHBC is a parallelogram. BC = AG because ABCG is a parallelogram. So AH = AG. AD is perpendicular to HG. So AD is simultaneously the perpendicular bisector of HG and the altitude from A to BC.
AC = BH because AHBC is a parallelogram. AC = BI because ABIC is a parallelogram. So BH = BI. BE is perpendicular to HI. So BE is simultaneously the perpendicular bisector of HI and the altitude from B to AC.
AB = CI because ABIC is a parallelogram. AB = CG because ABCG is a parallelogram. So CI = CG. CF is perpendicular to GI. So CF is simultaneously the perpendicular bisector of GI and the altitude from C to AB.
Since the perpendicular bisectors are concurrent, as proven in Theorem 3.3.12, the altitudes are also concurrent.
