Solving Problems with Similar Triangles

1. What are Similar Triangles?

In geometry, "similar" has a precise meaning. It's not just about looking alike. Two triangles are similar if their corresponding angles are equal and their corresponding sides are proportional.

Think about a photograph and an enlargement of it. The shapes are identical, but the sizes are different. Every part of the enlarged photo has been scaled up by the same factor. This is the core idea of similarity.

Let's get a formal definition.

As you've just read, if we say $△ABC\sim △PQR$, the order of the letters is crucial. It tells us which parts correspond:

• Corresponding Angles are Equal: $\angle A=\angle P$, $\angle B=\angle Q$, and $\angle C=\angle R$.
• Corresponding Sides are Proportional: The ratio of the lengths of corresponding sides is constant. This constant is often called the scale factor. $$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=k(\text{where }k\text{ is the scale factor})$$

2. The Criteria for Similarity

Thankfully, we don't need to check all six conditions (three angles and three sides) every time. There are three key shortcuts, or criteria, to prove that two triangles are similar. Mastering these is essential for quickly identifying similar triangles in exam problems.

Let's summarize these powerful criteria:

1. Angle-Angle (AA) Similarity: This is the most common and useful criterion for CAT problems. If two angles of one triangle are equal to two corresponding angles of another triangle, they are similar. (The third angles must also be equal, since the sum of angles in any triangle is 180°).

   • Typical CAT Scenario: Look for parallel lines, as they create equal corresponding or alternate angles, which often leads to AA similarity.

2. Side-Side-Side (SSS) Similarity: If the lengths of the three corresponding sides of two triangles are all in the same proportion, the triangles are similar.

   • Example: A triangle with sides 3, 4, 5 is similar to a triangle with sides 6, 8, 10 because the ratio is constant (1:2).

3. Side-Angle-Side (SAS) Similarity: If two sides of one triangle are proportional to two corresponding sides of another triangle, and the included angles (the angles between those two sides) are equal, then the triangles are similar.

   • Key detail: The equal angle must be between the proportional sides.

3. Properties and Ratios: The "So What?" of Similarity

Once you've established that two triangles are similar, you unlock a set of powerful relationships regarding their sides, perimeters, areas, and other corresponding lengths. This is where the problem-solving power lies.

Let's assume $△ABC\sim △PQR$ and the ratio of their corresponding sides is $\frac{AB}{PQ}=k$.

• Ratio of Perimeters: The ratio of their perimeters is the same as the ratio of their sides.

  $$\frac{\text{Perimeter}(△ABC)}{\text{Perimeter}(△PQR)}=k$$

• Ratio of Areas: This is a critical rule. The ratio of their areas is the square of the ratio of their corresponding sides.

  $$\frac{\text{Area}(△ABC)}{\text{Area}(△PQR)}={\left(\frac{AB}{PQ}\right)}^2=k^2$$

• Ratio of Corresponding Altitudes, Medians, etc.: The ratio of any corresponding linear elements (like altitudes, medians, angle bisectors) is also the same as the ratio of their sides.

Let's formally review these properties.

4. Application: Solving CAT-Style Problems

Now, let's see how these concepts are applied in practice. The following examples from the resource are typical of the logic you'll need for the CAT.

Analysis of Question 2: The Parallel Line Pattern

This is a fundamental pattern. A line drawn parallel to one side of a triangle creates a smaller, similar triangle at the top.

• Problem: In $△ABC$, $DE\parallel BC$. Given $DE=6$, $BC=8$, and Area($△ADE$) = 90, find Area($△ABC$).
• Step 1: Prove Similarity. Since $DE\parallel BC$, we know $\angle ADE=\angle ABC$ and $\angle AED=\angle ACB$ (corresponding angles). $\angle A$ is common to both $△ADE$ and $△ABC$. Therefore, by AA similarity, $△ADE\sim △ABC$.
• Step 2: Apply Area Ratio Theorem. $$\frac{\text{Area}(△ADE)}{\text{Area}(△ABC)}={\left(\frac{DE}{BC}\right)}^2$$
• Step 3: Solve. $$\frac{90}{\text{Area}(△ABC)}={\left(\frac{6}{8}\right)}^2={\left(\frac{3}{4}\right)}^2=\frac{9}{16}$$ $$\text{Area}(△ABC)=90\times \frac{16}{9}=10\times 16=160{\text{ cm}}^2$$

Analysis of Question 5: The Equal Area Division

This is a more advanced problem that beautifully combines the area theorem with a bit of algebra.

• Problem: In $△ABC$, $DE\parallel BC$. DE divides the triangle into two parts of equal area. Find the ratio AD : BD.
• Step 1: Set up the Area Relationship. We are told Area($△ADE$) = Area(trapezium BCED). This means Area($△ABC$) = Area($△ADE$) + Area(BCED) = 2 $\times$ Area($△ADE$).
  So, $\frac{\text{Area}(△ADE)}{\text{Area}(△ABC)}=\frac{1}{2}$.
• Step 2: Use Similarity to find the Side Ratio. We already know $△ADE\sim △ABC$. $$\frac{\text{Area}(△ADE)}{\text{Area}(△ABC)}={\left(\frac{AD}{AB}\right)}^2$$ $$\frac{1}{2}={\left(\frac{AD}{AB}\right)}^2⟹\frac{AD}{AB}=\frac{1}{\sqrt{2}}$$
• Step 3: Use Algebra to find the required ratio. We need AD : BD. From the previous step, we have $AB=\sqrt{2}\times AD$. Since $AB=AD+BD$, we can substitute: $$AD+BD=\sqrt{2}\times AD$$ $$BD=(\sqrt{2}-1)AD$$ $$\frac{AD}{BD}=\frac{1}{\sqrt{2}-1}$$ Thus, the ratio AD : BD is $1:(\sqrt{2}-1)$.

Conclusion

Today we've unpacked the concept of triangle similarity, a cornerstone of geometry for the CAT. By understanding how to identify similar triangles and apply their properties, you can solve a wide range of problems efficiently.

Key Takeaways:

• Similarity Conditions: You can prove triangles are similar using AA, SSS, or SAS criteria. The AA criterion is the most frequent in exam questions.
• Ratio of Sides vs. Ratio of Areas: This is the most critical distinction. If the ratio of corresponding sides is $k$, the ratio of perimeters is also $k$, but the ratio of areas is $k^2$.
• Common Pattern: A line parallel to the base of a triangle creates a smaller, similar triangle. This pattern is the basis for many problems.

Preview of the Next Lesson:
We will now move from triangles to another fundamental shape: circles. In the next lesson, we will solve problems using the angle, chord, and tangent properties of circles. These properties have their own unique set of rules that are frequently tested on the CAT.