KindiakMath

Similar Triangles

Having discussed a great deal about congruent triangles, we now turn our attention to a relaxation of congruence—similarities. Similarities are, no pun intended, similar ideas to congruence, but far more useful in helping us use small-scale items to represent large-scale items, such as maps and scale models.

Lemma 1. If \Delta ABC \equiv \Delta PQR, then

\displaystyle \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = 1.

Definition 1 (SSS Similarity Definition). We say that the triangles \Delta ABC, \Delta PQR are similar, denoted \Delta ABC \sim \Delta PQR, if there exists some positive constant k such that

\displaystyle \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = k.

Therefore, congruence becomes a special case of similarity via the special case k = 1.

Since congruent triangles have angles that match, do similar triangles have angles match? Not only do they match, but that they have to match.

Lemma 2. We have \Delta ABC \sim \Delta PQR if and only if their angles match:

\angle ABC = \angle PQR,\quad \angle BCA = \angle QRP,\quad \angle BAC = \angle QPR.

Proof Sketch. Suppose without loss of generality that AB < PQ. For setup, translate and rotate \Delta PQR where necessary so that A = P and B lies on PQ.

In the direction (\Rightarrow), suppose \Delta ABC \sim \Delta PQR. Then

\displaystyle \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR}.

A technical proof by contradiction shows that C must lie on PR. By the intercept theorem, BC \parallel QR. Using corresponding angles,

\angle ABC = \angle PQR,\quad \angle BCA = \angle QRP.

Since the angles in a triangle sum to 180^\circ, \angle BAC = \angle QPR.

In the direction (\Leftarrow), since \angle BAC = \angle QPR, C must lie on PR. Since

\angle ABC = \angle PQR,\quad \angle BCA = \angle QRP,

using corresponding angles, BC \parallel QR. By the intercept theorem again,

\displaystyle \frac{AB}{PQ} = \frac{AC}{PR} = \frac{BC}{QR},

yielding \Delta ABC \sim \Delta PQR, as required.

Theorem 1 (AA Similarity Criterion). We have \Delta ABC \sim \Delta PQR if and only if at least two of the following equalities hold:

\angle ABC = \angle PQR,\quad \angle BCA = \angle QRP,\quad \angle BAC = \angle QPR.

Proof. It suffices to prove the direction (\Leftarrow). Since angles in a triangle sum to 180^\circ, we get all three equalities if we know at least two of them hold. By Lemma 2, the result holds.

Theorem 2 (SAS Similarity Criterion). We have \Delta ABC \sim \Delta PQR if and only if

\displaystyle \angle ABC = \angle PQR,\quad \frac{AB}{PQ} = \frac{BC}{QR}.

Proof Sketch. It suffices to prove the direction (\Leftarrow). Since \angle ABC = \angle PQR, if A lies on PQ, then C must lie on QR.

By the intercept theorem,

\displaystyle \frac{AB}{PQ} = \frac{BC}{QR} \quad \Rightarrow \quad AC \parallel PQ.

Using corresponding angles, \angle BAC = \angle QPR. By the AA Similarity Criterion, \Delta ABC \sim \Delta PQR.

Similar triangles are used all the time in establishing interesting result in plane geometry.

Example 1 (Tangent-Secant Theorem). In the diagram below, \angle ABC = \angle BDC (this is called the alternate segment theorem).

Show that AB^2 = AC \cdot AD, known as the tangent-secant theorem.

Solution. Since \angle ABC = \angle BDC = \angle BDA and \angle BAC = \angle DBA, by the AA Similarity Criterion, \Delta ABC \sim \Delta BDA. In particular,

\displaystyle \frac{AB}{BD} = \frac{AC}{BA} = \frac{AC}{AB} \quad \Rightarrow \quad AB^2 = AC \cdot AD.

Example 2. In the diagram below, A,B,C,D are distinct points that lie on a circle.

Given that K is the intersection of AC and BD, show that

AK \cdot CK = BK \cdot DK.

Solution. For the first claim, since angles in the same segment are equal (i.e. the butterfly theorem), \angle ABK = \angle DCK. Since vertically opposite angles are equal, \angle AKB = \angle DKC. By the AA Similarity Criterion, \Delta ABK \sim \Delta DCK. In particular,

\displaystyle \frac{ AK }{ DK } = \frac{ BK }{ CK } \quad \Rightarrow \quad AK \cdot CK = BK \cdot DK.

Okay, let’s return to earth one more time.

Definition 2. Denote the scale 1 : N to mean that 1 unit of length in some representation (e.g. a map, scale model, etc) represents N units of length in real life.

Example 3. Suppose 1 cm on a map represents 5 km in real life. Determine the scale in terms of 1 : N.

Solution. Since 1\, \text m = 100\, \text{cm},

\begin{aligned} 5\, \text{km} &= (5 \times 10^3) \times 1\, \text{m} \\ &= (5 \times 10^3) \times 100\, \text{cm} \\ &= 500\, 000\, \text{cm}. \end{aligned}

Hence, the required scale is

\begin{aligned}1\, \text{cm} : 5\, \text{km} &= 1\, \text{cm} : 500\, 000\, \text{cm} = 1 : 500\, 000.\end{aligned}

Lemma 2. Suppose we have a scale of 1 : N. Then the area scale is given by 1 : N^2. That is, 1 square unit of area represents N^2 square units of area in real life.

Proof Sketch. We can use the Riemann integration approach of approximating objects using a combination of squares. Since a square with length and height 1 represents a real-life square with length and height N, the real-life square has area N \times N = N^2.

Lemma 3. Suppose we have a scale of 1 : N. Then the volume scale is given by 1 : N^3. That is, 1 cubic unit of area represents N^3 cubic units of area in real life.

Proof Sketch. Follow the idea in Lemma 2.

Likewise, using triangles to approximate shapes, two objects are similar if they share the same shape, and thus share some kind of similarity ratio in Definition 1. Therefore, the results following Definition 2 hold, allowing us to use scale models to represent real-life objects. From these ideas we obtain products like world globe maps and merchandise of varying sizes.

To further leverage these similarity properties, it helps for us to have simple formulas for well-known shapes. While we cannot prove them all in O-Level mathematics, we can state them and relegate their proofs to integral calculus. We explore these formulas in the next post.

—Joel Kindiak, 4 Nov 25, 1515H

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