Section 3.2 Similarity

Suppose we have \(\triangle ABC\text{.}\) Since point B is a point outside of the line \(\overline{AC}\text{,}\) we can construct a line going through B that is parallel to \(\overline{AC}\) by the Playfair Axiom. By doing so, we have constructed angles \(\angle \alpha\text{,}\) \(\angle \beta\text{.}\)

Since \(\overline{AC} \parallel L\text{,}\) the lines \(\overline{AB}\) and \(\overline{CB}\) are the transversals of the parallel lines, thus by the Alternate Interior Angle Converse Theorem, \(\angle\alpha \cong \angle ACB\) and \(\angle\beta \cong \angle CAB\text{.}\) By the Supplementary Postulate, \(m\angle\alpha + m\angle\beta + m\angle ABC = \pi\text{.}\) Thus, by congruence, \(m\angle ACB + m\angle CAB + m\angle ABC = \pi\text{.}\)

Suppose we have \(\triangle ABC\) with \(\overline{AB}\) extended to a point D, forming the exterior angle \(\angle ABD\text{.}\) By Playfair’s Axiom, we can construct a line L on point B that is parallel to \(\overline{AC}\) and in so doing, constructed angles \(\angle\alpha\text{,}\) \(\angle\beta\text{,}\) \(\angle\phi\text{.}\) By the Angle Addition Postulate, \(m\angle ABD = m\angle\beta + m\angle\phi\text{.}\) Since \(\overline{AC} \parallel L\text{,}\) \(\overline{AB}\) and \(\overline{CB}\) forms a transversal of the two parallel lines giving us that \(\angle\beta \cong \angle CAB\) and \(\angle\alpha \cong \angle ACB\) with \(\angle\alpha \cong \angle\phi\) by vertical angles, making \(\angle\phi \cong \angle ACB\text{.}\) Thus, \(m\angle ABD = m\angle CAB + m\angle ACB\text{.}\)

Establish the four parallel lines of our parallelogram, then transverse both pairs of parallels at either pair of opposing vertices. By the Converse of the Alternate Interior Angles Theorem, we know that \(\alpha\cong\beta\) and \(\gamma\cong\sigma\text{.}\) As the segment of the transverse in the interior of the parallelogram is congruent to itself, the parallelogram is constructed out of two congruent triangles, so \(\overline{AD}\cong\overline{BC}\) and \(\overline{AB}\cong\overline{CD}\text{.}\)

Consider three parallel lines \(l_1, l_2\text{,}\) and \(l_3\) intersected by a transversal \(t_1\) such that \(t_1\) is divided into two congruent sections. If \(t_1\) is perpendicular to the parallel lines, then the shortest distance between the neighboring lines is equal. Now consider the case of \(t_1\) not being perpendicular. Then using alternate interior angles, the angle formed by \(t_1\) and \(l_1\) is equal to the angle formed by \(t_1\) and \(l_3\text{.}\) We can draw the perpendicular from the intersection of \(t_1\) and \(l_2\) that spans from \(l_1\) to \(l_3\text{.}\) There now exist two triangles, one between \(l_1\) and \(l_2\) and one between \(l_2\) and \(l_3\text{.}\) The hypotenuse of each is equal as given, and the two corresponding angles are equal as shown with alternate interior angles and the definition of perpendicular. Hence these two triangles are congruent. Considering corresponding parts of congruent triangles are congruent, the two sets of parallels \(l_1 l_2\) and \(l_2 l_3\) are equidistant. Now consider some other transversal \(t_2\text{.}\) Draw the perpendicular from \(l_1\) to \(l_3\) through the \(t_2 l_2\) intersection. In accordance with the parallel postulate, these perpendicular segments are congruent. As they are perpendicular, each segment has a right angle on one end, and the other end is shown to be congruent through the vertical angles. Hence there are congruent triangles once again. Considering corresponding parts of congruent triangles are congruent, this arbitrary transversal is divided into congruent segments by the three parallel lines.

Assume there are three arbitrary parallel lines with a transversal that is divided into congruent segments. We will call the parallel lines \(\ell_1\text{,}\) \(\ell_2\text{,}\) and \(\ell_3\text{;}\) and the transversal will be t. This means that \(\overline{BC} \cong \overline{CD}\text{.}\) There exists an arbitrary transversal (T_1) that crosses the parallel lines. Using Playfair’s axiom, we can construct two lines parallel to \(T_1\) that intersects at point B and point C respectively, we will call them \(T_2\) and \(T_3\text{.}\) Using our corresponding angles lemma, we find that \(\angle CDG \cong \angle BCH\) and \(\angle DGC \cong \angle CHB\text{.}\) Then, using the Angle-Angle-Side and the congruence of components, we know \(\overline{CG} \cong \overline{BH}\text{.}\) Finally, we know that the parallel lines and the three transversals form quadrilaterals by definition. Using Theorem 3.2.5, we know that opposite sides of a parallelogram are congruent. Therefore, \(\overline{CG} \cong \overline{HI} \cong \overline{FJ}\) and \(\overline{BH} \cong \overline{EJ}\text{.}\) \(T_1\) is divided into congruent segments and any transversal can similarly be divided into congruent segments.

Suppose we have \(\triangle ABC\text{,}\) by the Ruler’s Postulate, we can construct a point D on the line \(\overline{AB}\) such that \(A-D-B\text{.}\) By Playfair’s Postulate, we can construct a line parallel to \(\overline{AC}\) and by Pasch’s Axiom this line will intersect \(\overline{BC}\) at a point E such that \(B-E-C\text{.}\) By Axiom 1, we can construct the lines \(\overline{DC}\) and \(\overline{EA}\text{,}\) thus constructing the triangles \(\triangle CDE\) and \(\triangle AED\text{.}\) From Checkpoint 3.2.10, we have \(\frac{A(\triangle DEB)}{A(\triangle AED)}\) = \(\frac{\frac{1}{2}\lvert\overline {FE}\rvert \lvert\overline{DB}\rvert}{\frac{1}{2} \lvert\overline{FE}\rvert \lvert\overline{AD}\rvert}\) = \(\frac{\lvert\overline{DB}\rvert}{\lvert\overline{AD}\rvert}\) and \(\frac{A(\triangle DEB)}{A(\triangle CDE)}\) = \(\frac{\frac{1}{2}\lvert\overline {DG}\rvert \lvert\overline{BE}\rvert}{\frac{1}{2} \lvert\overline{DG}\rvert \lvert\overline{EC}\rvert}\) = \(\frac{\lvert\overline{BE}\rvert}{\lvert\overline{EC}\rvert}\) and we have proven that \(A(\triangle AED) = A(\triangle CDE)\text{.}\) So, \(\frac{\lvert\overline{DB}\rvert}{\lvert\overline{AD}\rvert} = \frac{A(\triangle DEB)}{A(\triangle AED)} = \frac{A(\triangle DEB)}{A(\triangle CDE)} = \frac{\lvert\overline{BE}\rvert}{\lvert\overline{EC}\rvert}\text{.}\) Thus, we have \(\frac{\lvert\overline{DB}\rvert}{\lvert\overline{AD}\rvert} = \frac{\lvert\overline{BE}\rvert}{\lvert\overline{EC}\rvert}\) being proportional.

Assume there is an arbitrary triangle \(\triangle ABC\text{.}\) We will construct a line parallel to \(\overline{AC}\) using Playfair’s axiom with points D and E that lie on the line. Similarly, we will construct an additional line parallel to \(\overline{AB}\) with points E and F that lie on the line. By lemma 3.2.12, we know that \(\overline{AB}\) and \(\overline{BC}\) are divided proportionally. Using the corresponding angle lemma, we find that \(\angle BAC\cong\angle BDE\text{,}\) \(\angle BCA\cong\angle BED\text{,}\) and \(\angle ABC\cong\angle DBE\text{.}\) This means that \(\triangle DBE\) and \(\triangle FEC\) are similar triangles. We know that the points A, D, E, F form a parallelogram because both opposing sides are parallel. By Theorem 3.2.5, \(|DE|=|AF|\text{.}\) Using our known similar triangles, \(|AF|/|AC|=|BE|/|BC|\text{,}\) Then, by replacing \(|AF|\text{,}\) we find that \(|DE/|AC|=|BE|/|BC|\) which is what we wanted to show that \(\overline{AC}\) is divided proportionally. Therefore, the triangle is divided into two similar triangles where \(\triangle ABC\) is similar to \(\triangle BDE\)

By the SAS Postulate, \(\triangle AE'F'\cong\triangle DEF\text{.}\) Extend \(\overline{E'F'}\) to get vertical angles.

We see that we have two pairs of congruent alternate interior angles. By the Alternate Interior Angles Theorem (3.1.2), \(\overline{E'F'}\parallel\overline{BC}\text{.}\) So, by Corollary 3.2.13, \(\triangle ABC\) is similar to \(\triangle AE'F'\text{.}\) But \(\triangle AE'F'\cong\triangle DEF\text{.}\) So, \(\triangle ABC\) is similar to \(\triangle DEF\text{.}\)

If instead \(A-C-F'\text{,}\) extend \(\overline{BC}\) to get that \(\overline{E'F'}\parallel\overline{BC}\) by 3.1.2.

( \(\Leftarrow\) ) Suppose \(\triangle ABC\) and \(\triangle DEF\) are similar. Then, by definition, their corresponding angles are congruent. So, they share 3 congruent angles.