Chapter 6 Triangles (Concepts)

Congruent Figures and Similar Figures

In geometry, we often compare figures to understand their relationships. Two fundamental ways to compare figures are by checking for congruence and similarity. Congruence is about being identical, while similarity is about having the same shape, regardless of size.

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Congruent Figures

Two geometric figures are said to be congruent if they are exact copies of each other. This means they have the same shape and the same size. If you could "pick up" one figure, you could place it perfectly on top of the other through a series of slides (translations), turns (rotations), and flips (reflections).

Key Characteristics of Congruent Figures:

• All corresponding angles are equal.
• All corresponding sides are equal in length.

The symbol for congruence is $\cong$. If triangle ABC is congruent to triangle PQR, we write $\triangle ABC \cong \triangle PQR$.

Examples:

• Two circles with a radius of 5 cm are congruent.
• Two squares with a side length of 10 cm are congruent.
• Two line segments of length 7 cm are congruent.

Think of congruence as two objects being perfect duplicates, like two identical coins from the same mint.

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Similar Figures

Two geometric figures are similar if they have the same shape, but not necessarily the same size. One figure is essentially a scaled-up (enlarged) or scaled-down (reduced) version of the other. The transformation that relates them is called a dilation, or scaling.

Examples:

• Any two circles are similar to each other, regardless of their radii.
• Any two squares are similar to each other.
• Any two equilateral triangles are similar.
• A toy car and the real car it models are similar.

Conditions for Similarity in Polygons

For two polygons (like quadrilaterals, pentagons, etc.) to be similar, they must satisfy both of the following conditions:

1. All corresponding angles must be equal. This ensures they have the same shape.
2. All corresponding sides must be in the same ratio (or proportional). This ensures that one is a consistent scaling of the other.

This constant ratio of corresponding sides is called the scale factor or the ratio of similarity.

The symbol for similarity is $\sim$. If polygon ABCD is similar to polygon PQRS, we write $ABCD \sim PQRS$.

This means:

Condition 1 (Equal Angles):

$\angle A = \angle P, \quad \angle B = \angle Q, \quad \angle C = \angle R, \quad \angle D = \angle S$

Condition 2 (Proportional Sides):

Important: It's not enough for just one condition to be true. For example, a square and a non-square rectangle both have all angles equal to 90°, but their sides are not proportional, so they are not similar. Similarly, a square and a rhombus can have proportional sides, but their angles are not equal, so they are not similar.

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Relationship Between Congruence and Similarity

The two concepts are closely related. Think of congruence as a very special case of similarity.

• All congruent figures are similar. If two figures are congruent, they have the same shape (equal angles) and the same size (equal sides). Since their sides are equal, the ratio of corresponding sides is 1. This means they satisfy both conditions for similarity, with a scale factor of 1.
• Not all similar figures are congruent. Similar figures have the same shape, but they might have different sizes. If the scale factor between two similar figures is anything other than 1, their side lengths are different, so they cannot be congruent.

In short: Congruence = Similarity (with a scale factor of 1) + Same Size.

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Basic Proportionality Theorem (Thales Theorem)

The Basic Proportionality Theorem (BPT), also known as Thales's Theorem, is a cornerstone of geometry that describes a powerful relationship within a triangle. It establishes that if a line is drawn parallel to one side of a triangle, it will always divide the other two sides proportionally.

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Statement of the Basic Proportionality Theorem (BPT)

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Proof of BPT

Given: A triangle $\triangle ABC$ in which a line DE is drawn parallel to the side BC, intersecting the sides AB and AC at points D and E respectively.

To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$

Construction: Join BE and CD. Draw a perpendicular line from E to the side AB, let's call it $EM \perp AB$. Similarly, draw a perpendicular line from D to the side AC, let's call it $DN \perp AC$.

Proof:

We will use the formula for the area of a triangle: Area $= \frac{1}{2} \times \text{base} \times \text{height}$.

Part 1: Relating the segments of side AB.

Consider $\triangle ADE$. If we take AD as the base, its corresponding height is EM.

Area($\triangle ADE$) $= \frac{1}{2} \times AD \times EM$.

Now consider the obtuse triangle $\triangle DBE$. If we take DB as the base, its height is also EM (the perpendicular distance from E to the line containing DB).

Area($\triangle DBE$) $= \frac{1}{2} \times DB \times EM$.

Let's take the ratio of their areas:

Part 2: Relating the segments of side AC.

Consider $\triangle ADE$ again. This time, let's take AE as the base. Its corresponding height is DN.

Area($\triangle ADE$) $= \frac{1}{2} \times AE \times DN$.

Now consider the obtuse triangle $\triangle ECD$. If we take EC as the base, its height is also DN.

Area($\triangle ECD$) $= \frac{1}{2} \times EC \times DN$.

Let's take the ratio of their areas:

Part 3: Connecting the two parts.

Now, observe the two triangles $\triangle DBE$ and $\triangle ECD$.

• They lie on the same base, DE.
• They are between the same parallel lines, DE and BC (since it is given that $DE \parallel BC$).

A theorem from earlier classes states that triangles on the same base and between the same parallels are equal in area.

Therefore, Area($\triangle DBE$) = Area($\triangle ECD$).

Now look at equations (1) and (2). Their numerators are the same (Area($\triangle ADE$)), and we have just shown that their denominators are also equal. If two fractions have equal numerators and equal denominators, the fractions themselves must be equal.

From (1) and (2), this means:

This completes the proof of the Basic Proportionality Theorem.

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Converse of the Basic Proportionality Theorem

Given: A triangle $\triangle ABC$ and a line DE intersecting sides AB and AC at D and E respectively, such that $\frac{AD}{DB} = \frac{AE}{EC}$.

To Prove: $DE \parallel BC$.

Proof by Contradiction:

Step 1: Make an assumption.

Let us assume that the line DE is not parallel to BC. If this is true, then there must be another line passing through D that is parallel to BC. Let's draw a line DE' such that $DE' \parallel BC$, where E' is a point on the side AC.

Step 2: Apply the original BPT.

Since we have constructed $DE' \parallel BC$, we can apply the Basic Proportionality Theorem (Theorem 6.1) to $\triangle ABC$. According to BPT:

Step 3: Use the given information.

It is given in the problem statement that the line DE divides the sides in the same ratio:

Step 4: Equate the results.

From equations (i) and (ii), since both right-hand sides are equal to $\frac{AD}{DB}$, they must be equal to each other:

Step 5: Manipulate the equation to find a contradiction.

Add 1 to both sides of the equation:

Find a common denominator for each side:

From the figure, we can see that $AE' + E'C = AC$ and $AE + EC = AC$. Substitute this into the equation:

Since the numerators are the same (and not zero), the denominators must be equal:

This result implies that the distance from E' to C is the same as the distance from E to C. Since both E and E' lie on the same line segment AC, this is only possible if the points E and E' are the same point, i.e., they coincide.

Step 6: Conclude the proof.

If E and E' are the same point, then the line DE is the same as the line DE'. We started by constructing DE' to be parallel to BC. Therefore, DE must also be parallel to BC.

This contradicts our initial assumption that DE was not parallel to BC. Thus, our assumption was wrong.

Hence, the line DE must be parallel to BC.

(Hence Proved)

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Similarity of Triangles and Their Criteria of Similarity

In the previous section, we introduced the concept of similar figures as figures having the same shape but not necessarily the same size. For polygons with the same number of sides to be similar, their corresponding angles must be equal, and their corresponding sides must be proportional. When we apply this concept to triangles, these conditions simplify, leading to specific criteria for proving triangle similarity.

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Similarity of Triangles

Two triangles are said to be similar if:

1. Their corresponding angles are equal.
2. Their corresponding sides are in the same ratio (proportional).

If $\triangle \text{ABC}$ is similar to $\triangle \text{PQR}$, we write $\triangle \text{ABC} \sim \triangle \text{PQR}$. The order of the vertices is important; it indicates the correspondence between the vertices and hence the corresponding angles and sides.

If $\triangle \text{ABC} \sim \triangle \text{PQR}$, then:

• Corresponding angles are equal: $\angle \text{A} = \angle \text{P}$, $\angle \text{B} = \angle \text{Q}$, $\angle \text{C} = \angle \text{R}$.
• Corresponding sides are proportional: $\frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{CA}}{\text{RP}} = k$ (where $k$ is the scale factor).

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Criteria for Similarity of Triangles

Just as with congruent triangles, we do not need to check all six conditions (three angles and three side ratios) to determine if two triangles are similar. Specific combinations of conditions are sufficient to prove similarity. These are called the criteria for similarity of triangles.

1. AAA Similarity Criterion (Angle-Angle-Angle)

If in $\triangle \text{ABC}$ and $\triangle \text{PQR}$:

Then $\triangle \text{ABC} \sim \triangle \text{PQR}$.

Note: The sum of angles in any triangle is $180^\circ$. Therefore, if two angles of one triangle are respectively equal to two angles of another triangle, the third angles must automatically be equal. This leads to a simplified version of the AAA criterion, known as the AA Similarity Criterion.

AA Similarity Criterion (Angle-Angle)

If in $\triangle \text{ABC}$ and $\triangle \text{PQR}$:

Then $\triangle \text{ABC} \sim \triangle \text{PQR}$.

This AA criterion is most commonly used in problems as it is simpler to apply.

2. SSS Similarity Criterion (Side-Side-Side)

If in $\triangle \text{ABC}$ and $\triangle \text{PQR}$:

Then $\triangle \text{ABC} \sim \triangle \text{PQR}$.

3. SAS Similarity Criterion (Side-Angle-Side)

If in $\triangle \text{ABC}$ and $\triangle \text{PQR}$:

Then $\triangle \text{ABC} \sim \triangle \text{PQR}$.

Note the crucial condition: the angle must be the one included between the two proportional sides.

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Similarity of Triangles formed by an Altitude to the Hypotenuse

A particularly important and elegant result in geometry arises when we draw an altitude from the vertex of the right angle in a right-angled triangle down to its hypotenuse. This single line segment divides the original triangle into two smaller triangles, and all three triangles—the original large one and the two smaller new ones—are similar to each other.

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The Theorem of Similarity in a Right Triangle

Given: A right-angled triangle $\triangle ABC$, with the right angle at B ($\angle B = 90^\circ$). An altitude BD is drawn from vertex B to the hypotenuse AC.

To Prove:

1. $\triangle ADB \sim \triangle ABC$
2. $\triangle BDC \sim \triangle ABC$
3. $\triangle ADB \sim \triangle BDC$

Proof Strategy: We will use the AA (Angle-Angle) Similarity Criterion for each part of the proof.

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Proof of the Theorem

Part 1: Proving $\triangle ADB \sim \triangle ABC$

We need to find two pairs of equal angles in $\triangle ADB$ and the original triangle $\triangle ABC$.

1. Angle 1 (Common Angle):

$\angle A = \angle A$

(It is a common angle to both triangles)

2. Angle 2 (Right Angles):

$\angle ADB = 90^\circ$

(Given that $BD \perp AC$)

$\angle ABC = 90^\circ$

(Given for $\triangle ABC$)

Therefore, $\angle ADB = \angle ABC$.

Since two angles of $\triangle ADB$ are equal to two corresponding angles of $\triangle ABC$, by the AA Similarity Criterion, we have:

$\triangle ADB \sim \triangle ABC$. (Part 1 is proved)

Part 2: Proving $\triangle BDC \sim \triangle ABC$

Similarly, we find two pairs of equal angles in $\triangle BDC$ and the original triangle $\triangle ABC$.

1. Angle 1 (Common Angle):

$\angle C = \angle C$

(It is a common angle to both triangles)

2. Angle 2 (Right Angles):

$\angle BDC = 90^\circ$

(Given that $BD \perp AC$)

$\angle ABC = 90^\circ$

(Given for $\triangle ABC$)

Therefore, $\angle BDC = \angle ABC$.

Since two angles of $\triangle BDC$ are equal to two corresponding angles of $\triangle ABC$, by the AA Similarity Criterion, we have:

$\triangle BDC \sim \triangle ABC$. (Part 2 is proved)

Part 3: Proving $\triangle ADB \sim \triangle BDC$

Now we prove that the two smaller triangles are similar to each other.

Method 1: Using Transitivity

From Part 1, we know $\triangle ADB \sim \triangle ABC$.

From Part 2, we know $\triangle BDC \sim \triangle ABC$.

Since both smaller triangles are similar to the same large triangle, they must be similar to each other. Therefore:

$\triangle ADB \sim \triangle BDC$. (Part 3 is proved)

Method 2: Direct Proof using AA Similarity

Let's find two pairs of equal angles in $\triangle ADB$ and $\triangle BDC$.

1. Angle 1 (Right Angles):

$\angle ADB = \angle BDC = 90^\circ$

(Given)

2. Angle 2 (Using Angle Sum Property):

In the large triangle $\triangle ABC$, since $\angle B = 90^\circ$, we know that $\angle A + \angle C = 90^\circ$. Let's call this ... (i)

Now, look at the smaller triangle $\triangle ADB$. It is also a right triangle ($\angle D = 90^\circ$), so its other two angles must add up to 90°:

$\angle A + \angle ABD = 90^\circ$

... (ii)

Comparing equations (i) and (ii):

$\angle A + \angle C = 90^\circ$

$\angle A + \angle ABD = 90^\circ$

This implies that $\angle C = \angle ABD$.

So, in $\triangle ADB$ and $\triangle BDC$, we have found two pairs of equal angles:

• $\angle ADB = \angle BDC$ (both 90°)
• $\angle ABD = \angle C$ (proved above)

Therefore, by the AA Similarity Criterion:

$\triangle ADB \sim \triangle BDC$. (Part 3 is proved)

This completes the proof of the entire theorem.

Areas of Similar Triangles

When two triangles are similar, not only are their sides proportional, but their areas are also related in a very specific and predictable way. This relationship is captured in a key theorem that connects the ratio of their areas to the ratio of their corresponding sides.

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Theorem on the Areas of Similar Triangles

Given: Two triangles, $\triangle ABC$ and $\triangle PQR$, such that $\triangle ABC \sim \triangle PQR$.

This similarity implies:

• Their corresponding angles are equal: $\angle A = \angle P, \angle B = \angle Q, \angle C = \angle R$.
• Their corresponding sides are proportional: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}$.

To Prove:

Construction: Draw altitudes from vertices A and P to their opposite sides. Let $AM \perp BC$ and $PN \perp QR$.

Proof:

Step 1: Write the ratio of the areas.

Using the formula Area $= \frac{1}{2} \times \text{base} \times \text{height}$:

Area($\triangle ABC$) $= \frac{1}{2} \times BC \times AM$

Area($\triangle PQR$) $= \frac{1}{2} \times QR \times PN$

The ratio of their areas is:

Step 2: Find the ratio of the altitudes.

Now, consider the smaller triangles $\triangle ABM$ and $\triangle PQN$.

• $\angle B = \angle Q$ (Since $\triangle ABC \sim \triangle PQR$)
• $\angle AMB = \angle PNQ = 90^\circ$ (By construction)

By the AA Similarity Criterion, $\triangle ABM \sim \triangle PQN$.

Since these smaller triangles are similar, their corresponding sides are proportional:

Step 3: Connect the ratios.

We know from the original similarity ($\triangle ABC \sim \triangle PQR$) that:

From equations (2) and (3), we can conclude that the ratio of the altitudes is the same as the ratio of the bases:

Step 4: Substitute and complete the proof.

Now, substitute the result from (4) back into our area ratio in equation (1):

Since $\frac{BC}{QR}$ is equal to the ratio of any other pair of corresponding sides (from equation 3), we can write the final result:

(Hence Proved)

This theorem also extends to other corresponding linear measurements, such as the ratio of medians, angle bisectors, and perimeters. The ratio of areas will be the square of the ratio of any of these corresponding lengths.

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Worked Examples

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Pythagoras Theorem

The Pythagoras Theorem is one of the most famous and foundational theorems in all of mathematics. It describes a simple yet profound relationship between the three sides of a right-angled triangle. It is a powerful tool used in construction, navigation, physics, and countless other fields.

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Statement of the Pythagoras Theorem

Given: A right-angled triangle $\triangle ABC$, with the right angle at vertex B ($\angle B = 90^\circ$). The side AC is the hypotenuse.

To Prove: $AB^2 + BC^2 = AC^2$

Proof using Similarity

This elegant proof relies on the theorem about the similarity of triangles formed by drawing an altitude to the hypotenuse.

Construction: Draw an altitude BD from the vertex B to the hypotenuse AC.

Proof:

We know from a previous theorem that drawing the altitude BD creates three similar triangles: $\triangle ADB \sim \triangle ABC$ and $\triangle BDC \sim \triangle ABC$.

Step 1: Use the similarity of $\triangle ADB$ and $\triangle ABC$.

Since $\triangle ADB \sim \triangle ABC$, the ratio of their corresponding sides is equal. Matching the corresponding vertices ($A \leftrightarrow A$, $D \leftrightarrow B$, $B \leftrightarrow C$):

By cross-multiplying, we get:

Step 2: Use the similarity of $\triangle BDC$ and $\triangle ABC$.

Since $\triangle BDC \sim \triangle ABC$, the ratio of their corresponding sides is equal. Matching the corresponding vertices ($C \leftrightarrow C$, $D \leftrightarrow B$, $B \leftrightarrow A$):

By cross-multiplying, we get:

Step 3: Add the two results.

Adding equation (1) and equation (2):

Factor out the common term AC on the right side:

From the figure, we can see that the segment AD and the segment DC together make up the entire hypotenuse AC. So, $AD + DC = AC$.

Substitute this back into the equation:

(Hence Proved)

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Converse of Pythagoras Theorem

Given: A triangle $\triangle ABC$ where the sides satisfy the relation $AB^2 + BC^2 = AC^2$.

To Prove: $\angle B = 90^\circ$.

Construction: Construct another triangle, $\triangle PQR$, such that it is a right-angled triangle at Q ($\angle Q = 90^\circ$), and its legs are equal to the legs of $\triangle ABC$. That is, $PQ = AB$ and $QR = BC$.

Proof:

Step 1: Apply Pythagoras Theorem to the constructed triangle.

In $\triangle PQR$, since $\angle Q = 90^\circ$, by Pythagoras Theorem:

Step 2: Use construction and given information.

By construction, we made $PQ = AB$ and $QR = BC$. Substitute these into equation (i):

We are given for $\triangle ABC$ that:

Step 3: Compare the sides.

From (ii) and (iii), we see that $PR^2 = AC^2$, which means $PR = AC$.

Step 4: Prove triangle congruence.

Now, compare $\triangle ABC$ and $\triangle PQR$:

• $AB = PQ$ (By construction)
• $BC = QR$ (By construction)
• $AC = PR$ (Proved above)

By the SSS (Side-Side-Side) Congruence Criterion, $\triangle ABC \cong \triangle PQR$.

Step 5: Conclude the proof.

Since the triangles are congruent, their corresponding angles must be equal. Therefore, $\angle B$ must be equal to $\angle Q$.

By construction, $\angle Q = 90^\circ$. So, $\angle B = 90^\circ$.

(Hence Proved)

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