Similarity (dilations, AA, proportional segments) Knowledge-Based Study

Study notes

1. What is a Dilation?

   A dilation is a non-rigid geometric transformation that alters the size of a figure without altering its shape. Every dilation is determined by two critical components:

   • Center of dilation $O$: A fixed reference point from which all points are stretched or compressed along rays.
   • Scale factor $k$: A non-zero real number that dictates the ratio of the image distance to the preimage distance relative to the center. For any point $P$ and its image $P'$, we have $OP' = |k| \cdot OP$.

   When $k > 1$, the transformation is an enlargement. When $0 < k < 1$, it is a reduction. Dilations map lines to parallel lines (or to themselves if passing through the center), preserve collinearity, and crucially, preserve angle measures.

   The Definition of Similarity

   Two geometric figures are defined as similar (denoted by the symbol $\sim$) if and only if one figure can be mapped onto the other through a sequence of rigid motions (translations, rotations, reflections) combined with at least one dilation.

   As a direct consequence of this definition, two polygons are similar if and only if:

   1. All corresponding angles are congruent: $\angle A \cong \angle A'$, $\angle B \cong \angle B'$, etc.
   2. All corresponding side lengths are proportional: $$\frac{A'B'}{AB} = \frac{B'C'}{BC} = \frac{C'D'}{CD} = k$$

   Establishing Similarity in Triangles

   While general polygons require congruence of all angles and proportionality of all sides, triangles have unique rigidity properties. The most fundamental and widely used criterion is the Angle-Angle (AA) Similarity Criterion.

   The Angle-Angle (AA) Criterion

   If two angles of one triangle are congruent to two angles of another triangle, then the two triangles are similar.

   Theorem Statement: If in $riangle ABC$ and $riangle DEF$, we have $\angle A \cong \angle D$ and $\angle B \cong \angle E$, then $$riangle ABC \sim riangle DEF$$

   Why is AA sufficient? Because the interior angles of any Euclidean triangle sum to $180^\circ$, if two pairs of angles are congruent, the third pair must also be congruent: $$\angle C = 180^\circ - (\angle A + \angle B) = 180^\circ - (\angle D + \angle E) = \angle F$$ This angle equality ensures that one triangle is an exact scaled dilation of the other, making side proportionality an inevitable mathematical consequence.

   Other Related Criteria

   • SAS (Side-Angle-Side) Similarity: If an angle of one triangle is congruent to an angle of another triangle, and the sides including those angles are proportional, the triangles are similar.
   • SSS (Side-Side-Side) Similarity: If the corresponding sides of two triangles are proportional, the triangles are similar.

   The Triangle Proportionality Theorem

   A central application of similarity involves parallel lines cutting across the sides of a triangle.

   Theorem: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides proportionally.

   Consider $riangle ABC$. Let point $D$ lie on side $AB$ and point $E$ lie on side $AC$ such that $DE \parallel BC$:

   • Because $DE \parallel BC$, the corresponding angles are congruent: $\angle ADE \cong \angle ABC$ and $\angle AED \cong \angle ACB$.
   • By the AA Similarity Criterion, $riangle ADE \sim riangle ABC$.
   • From triangle similarity, the ratio of corresponding sides is constant: $$\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$$
   • Decomposing $AB$ as $AD + DB$ and $AC$ as $AE + EC$ yields the split-segment ratio: $$\frac{AD}{DB} = \frac{AE}{EC}$$

   Converse of the Triangle Proportionality Theorem

   If a line divides two sides of a triangle proportionally, such that $\frac{AD}{DB} = \frac{AE}{EC}$, then the line is parallel to the third side ($DE \parallel BC$).

   Three Parallel Lines (Thales' Intercept Theorem)

   If three or more parallel lines intersect two transversals, they cut off the transversals proportionally. If lines $l_1 \parallel l_2 \parallel l_3$ intersect transversal $t_1$ at points $A, B, C$ and transversal $t_2$ at points $D, E, F$, then: $$\frac{AB}{BC} = \frac{DE}{EF}$$

   Application 1: Solving for Missing Sides Using AA Similarity

   Problem: In $riangle PQR$, a segment $ST$ is drawn with $S$ on $PQ$ and $T$ on $PR$ such that $\angle PST \cong \angle R$. If $PS = 4$, $SQ = 5$, and $PT = 3$, find the length of $TR$.

   Solution Strategy:

   1. Identify Common and Congruent Angles: Both $riangle PST$ and $riangle PRQ$ share angle $\angle P$. It is given that $\angle PST \cong \angle PRQ$.
   2. Establish Similarity: By the AA criterion, $riangle PST \sim riangle PRQ$ (note the careful ordering of corresponding vertices: $P \leftrightarrow P$, $S \leftrightarrow R$, $T \leftrightarrow Q$).
   3. Set Up the Proportion: $$\frac{PS}{PR} = \frac{PT}{PQ}$$
   4. Substitute Known Values: Note that $PQ = PS + SQ = 4 + 5 = 9$. Let $TR = x$, so $PR = PT + TR = 3 + x$. $$\frac{4}{3 + x} = \frac{3}{9}$$
   5. Solve for $x$: Since $\frac{3}{9} = \frac{1}{3}$, we have: $$4 \cdot 3 = 1 \cdot (3 + x) \implies 12 = 3 + x \implies x = 9$$ Thus, $TR = 9$.

   Application 2: Using the Triangle Proportionality Theorem

   Problem: In $riangle XYZ$, line segment $MN$ is drawn parallel to $YZ$, with $M$ on $XY$ and $N$ on $XZ$. If $XM = 6$, $MY = 2x - 1$, $XN = 8$, and $NZ = 3x - 2$, determine the value of $x$.

   Solution:

   1. Since $MN \parallel YZ$, by the Triangle Proportionality Theorem: $$\frac{XM}{MY} = \frac{XN}{NZ}$$
   2. Substitute the given algebraic expressions: $$\frac{6}{2x - 1} = \frac{8}{3x - 2}$$
   3. Cross-multiply and solve the linear equation: $$6(3x - 2) = 8(2x - 1)$$ $$18x - 12 = 16x - 8$$ $$2x = 4 \implies x = 2$$