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An IBL Introduction to Geometries

Mark A. Fitch

Section 3.2 Similarity

Subsection 3.2.1 Preparation Theorems

Checkpoint 3.2.1.

For \(\triangle ABC\) construct line \(\ell\) such that \(\ell \parallel \overline{AC}\) and \(B\) is on \(\ell.\) What is the relationship between the three angles at \(B\) (smaller than a straight angle) to the angles of the triangle?
Let \(\triangle ABC\) exist and construct line \(l\) so \(l\) is parallel to \(\overline{AC}\) and \(B\) is on \(l.\)
The sum of the measures of angles \(d, b\) and \(e\) is equal to \(\pi\) because these angles together form a line.
Angle \(d\) is congruent with angle \(a\) and angle \(e\) is congruent with angle \(c\) from the alternate interior angle theorem.
Since the sum of the measures of angles \(d, b\) and \(e\) equal \(\pi,\) the sum of the measures of \(a, b\) and \(c\) also equal \(\pi\) because of the angle congruences we stated.
Consider a triangle \(\triangle ABC\) with angles \(\angle BAC = a\text{,}\) \(\angle ABC = b\text{,}\) and \(\angle BCA = c\text{,}\) where segments \(\overline{CA}, \overline{AB}\text{,}\) and \(\overline{BC}\) are extended to construct exterior angles \(a^\prime\text{,}\) \(b^\prime\text{,}\) and \(c^\prime\) respectively. We know the interior angles add to \(180^\circ\) by Theorem 3.2.2, so \(a + b + c = 180^\circ\text{.}\) Note \(a^\prime + a = 180^\circ\text{.}\) Also, \(a = 180 - b - c\text{,}\) so \(a^\prime + 180^\circ - b - c = 180^\circ\) implies \(a^\prime = b + c\text{.}\) Because this can be done without loss of generality, the proof is shown.

Definition 3.2.4. Parallelogram.

A quadrilateral is a parallelogram if and only if both opposing pairs of sides are parallel.
\(\overline{AB}\) and \(\overline{CD}\) are parallel, and \(\overline{AC}\) and \(\overline{BD}\) are parallel.
Construct line \(\overline{AD}.\) From AIA converse theorem, angle \(CAD\) and angle \(ADB\) are congruent, and angle \(BAD\) and angle \(ADC\) are congruent.
Now use the Angle-Side-Angle theorem: \(\triangle ADC\) and \(\triangle DAB\) are congruent.
Since the triangles are congruent, their corresponding sides are congruent. Thus, \(\overline{AC}\) and \(\overline{DB}\) are congruent, and \(\overline{AB}\) and \(\overline{DC}\) are congruent.
Suppose that parallel lines \(l\text{,}\) \(m\text{,}\) and \(n\) are intersected by transversal \(t\) at points \(A\text{,}\) \(B\text{,}\) and \(C\) respectively. Further suppose that \(|AB| = |BC|\text{.}\) Let \(u\) be any other transversal of the three initial lines with intersections \(D\text{,}\) \(E\text{,}\) and \(F\text{.}\)
Consider the right triangles \(\triangle AA^\prime B\) and \(\triangle BB^\prime C\) formed by dropping the foot of the perpendicular from \(A\) and \(B\) respectively. These triangles have two congruent angles and a congruent side, and are therefore congruent. (The angle congruence follows from alternate interior and vertical angles.) Thus, the line segments \(AA^\prime\) and \(BB^\prime\) are congruent.
By the equidistance axiom, lines \(l\) is everywhere equidistant from line \(m\text{,}\) and line \(m\) from line \(n\text{.}\) Since \(|AA^\prime|\) and \(|BB^\prime|\) are the distances between \(l\) and \(m\) and \(m\) and \(n\) respectivly, we now know that \(l\) is the same distance from \(m\) as \(m\) is from \(n\text{.}\)
Now consider the triangles formed by dropping the foot of the perpendicular from \(D\) to \(m\) and \(E\) to \(n\text{.}\) Then by the foregoing \(|DD^\prime| = |EE^\prime| = |AA^\prime| = |BB|\text{.}\) Thus triangles \(\triangle DD^\prime E\) and \(\triangle EE^\prime F\) are congruent since they have one congruent side and two congruent angles. Since corresponding parts of congruent triangles are congruent, \(\overline{DE} \cong \overline{EF}\text{.}\) Since \(u\) was an arbitrary transversal, the theorem follows.

Subsection 3.2.2 Explore Similarity Theorems

Definition 3.2.7. Altitude.

A line segment is an altitude if it connects a vertex of a triangle to the foot of the perpendicular on the opposite side.

Checkpoint 3.2.8.

Construct a triangle and enough parallel lines to divide one side of the triangle into four equal parts. Into how many parts do these lines divide the other sides?

Definition 3.2.9. Triangle Area.

The area of a triangle is equal to one half of the product of one side times the length of the altitude from the opposing vertex to that side.

Checkpoint 3.2.10.

Construct \(\triangle ABC\text{.}\) Construct \(\overline{DE}\) such that \(B-D-A,\) \(B-E-C\text{,}\) and \(\overline{DE} \parallel \overline{AC}.\)
(a)
Construct \(\overline{AE}\) and \(\overline{EF}\) such that \(F\) is the foot of the perpendicular from \(E.\) Reduce the ratio of the areas of \(\triangle DEB\) and \(\triangle AED.\)
Solution.
\(\overline{AE}\) and \(\overline{EF}\) are constructed. The area of a triangle is defined as "one half of the product of one side times the length of the altitude from the opposing vertex to that side." Thus, the ratio of the areas of \(\triangle DEB\) and \(\triangle AED\) is \(\frac{(\frac{1}{2}) \cdot |BD| \cdot |EF|}{(\frac{1}{2}) \cdot |AD| \cdot |EF|}.\) Note that both triangles share altitude \(|EF|.\) The ratio reduces to \(\frac{|BD|}{|AD|}.\)
(b)
Construct \(\overline{DC}\) and \(\overline{DG}\) such that \(G\) is the foot of the perpendicular from \(D.\) Reduce the ratio of the areas of \(\triangle DEB\) and \(\triangle CDE.\)
Solution.
\(\overline{DC}\) and \(\overline{DG}\) are constructed. By the same process as before, the ratio of the areas of \(\triangle DEB\) and \(\triangle CDE\) is \(\frac{(\frac{1}{2}) \cdot |BE| \cdot |DG|}{\frac{1}{2}) \cdot |CE| \cdot |DG|}.\) The triangles share altitude \(|DG|.\) The ratio reduces to \(\frac{|BE|}{|CE|}.\)
(c)
Prove area of \(\triangle AED\) is equal to the area of \(\triangle CDE.\)
Solution.
Construct a line from \(A\) to the foot of the perpendicular on \(\overline{DE}.\) Construct another line from \(C\) to the foot of the perpendicular on \(\overline{DE}.\) By point to line distance theorem, the shortest distance from \(A\) to \(\overline{DE}\) and \(C\) to \(\overline{DE}\) are the lines that we just constructed. Since \(\overline{AC}\) and \(\overline{DE}\) are parallel, the lines are everywhere equidistant. Thus, the lines we constructed are of equal length, call it \(m.\) The area of \(\triangle AED\) is \(\frac{\frac{1}{2} \cdot |DE| \cdot m}{|CE|}\) since \(|DE|\) is a side of the triangle and \(m\) is the length of the altitude from \(|DE|\) to the opposing vertex. The area of \(\triangle CDE\) is also \(\frac{\frac{1}{2} \cdot |DE| \cdot m}{|CE|}\) since \(|DE|\) is a side of the triangle and \(m\) is the length of the altitude from \(|DE|\) to the opposing vertex! Thus, the areas of \(\triangle AED\) and \(\triangle CDE\) are equal.

Subsection 3.2.3 Similarity Theorems

Definition 3.2.11. Similar Triangles.

Two triangles are similar if and only if corresponding angles are congruent and the ratio of corresponding sides is constant.
Consider a line parallel to one side of a triangle \(\triangle ABC\) and intersecting the other two sides, say, at points \(D\) and \(E\) on \(\overline{AB}\) and \(\overline{AC}\) respectively. We would like to show \(\frac{|\overline{AD}|}{|\overline{BD}|} = \frac{|\overline{AE}|}{|\overline{AC}|}\text{.}\) If we apply the steps from Checkpoint 3.2.10, we know that \(A(\triangle BED) = A(\triangle CDE)\text{.}\) This means \(\frac{1}{2}(|\overline{BD}|)(|\overline{EF}|) = \frac{1}{2}(|\overline{CE}|)(|\overline{GD}|)\text{.}\) Also, we know that \(A(\triangle ADE) = \frac{1}{2}(|\overline{EF}|)(|\overline{AD}|) = \frac{1}{2}(|\overline{GD}|)(|\overline{AE}|)\) which implies \(\frac{|\overline{EF}|}{|\overline{GD}|} = \frac{|\overline{AE}|}{|\overline{AD}|}\text{.}\) Therefore, because \((|\overline{BD}|)(|\overline{EF}|) = (|\overline{CE}|)(|\overline{GD}|)\) implies \(\frac{|\overline{EF}|}{|\overline{GD}|} = \frac{|\overline{CE}|}{|\overline{BD}|}\text{,}\) then \(\frac{|\overline{AE}|}{|\overline{AD}|} = \frac{|\overline{CE}|}{|\overline{BD}|}\) implies \(\frac{|\overline{AD}|}{|\overline{BD}|} = \frac{|\overline{AE}|}{|\overline{CE}|}\text{,}\) so the sides are divided proportionally.
Let \(\triangle ABC\) have \(\overline{DE} || \overline{AC}\) so \(B - D - A\) and \(B - E - C.\) Angle \(a\) and angle \(d\) are congruent, and angle \(c\) and angle \(e\) are congruent by AIA converse theorem and vertical angle definition. The two triangles share angle \(b.\) So, the corresponding angles of \(\triangle ABC\) and \(\triangle DBE\) are congruent.
Construct \(\overline{AE}\) and \(\overline{EF}\) so \(F\) is the foot of the perpendicular from \(E.\) Also construct \(|DC|\) and \(|DG|\) so \(G\) is the foot of the perpendicular from \(D.\) From Lemma 3.2.12, we know that \(\frac{|BD|}{|AB|} = \frac{|BE|}{|CE|}.\)
Construct \(\overline{DH}\) parallel to \(\overline{BC}\) so \(A - H - C.\) From Lemma 3.2.12, this line divides the sides \(|AB|\) and \(|AC|\) proportionally. So \(\frac{|BD|}{|AB|} = \frac{|CH|}{|AC|}.\) Remember that \(\frac{|BD|}{|AB|} = \frac{|BE|}{|CE|},\) so \(\frac{|BD|}{|AB|} = \frac{|BE|}{|CE|} = \frac{|CH|}{|AC|}.\) \(DHCE\) is a parallelogram by definition because \(\overline{DH} || \overline{CE}\) and \(\overline{DE} || \overline{CH}.\) By Theorem 3.2.5 \(|DE| = |CH|\) because they are opposing sides, so we can substitute the value in our earlier equation: \(\frac{|BD|}{|AB|} = \frac{|BE|}{|CE|} = \frac{|DE|}{|AC|}.\) By definition of similarity, because the corresponding angles of \(\triangle ABC\) and \(\triangle DBE\) are congruent and the ratio of corresponding sides is constant, \(\triangle ABC\) and \(\triangle DBE\) are similar.

Subsection 3.2.4 Extending Similarity

Checkpoint 3.2.15.

Construct a definition for similar quadrilaterals. Construct examples to show that your definition works.

Checkpoint 3.2.16.

Explain why similarity is not defined simply as "all angles are congruent."
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