Given: The definition of similar figures in geometry, which states that two figures are similar if they have the same shape but not necessarily the same size.

To Find: Two distinct examples of pairs of similar figures.

Theoretical Context: Two polygons are said to be similar if:
1. Their corresponding angles are equal.
2. Their corresponding sides are in the same ratio (proportion).

Visual Representation:

Step 1: Identifying Example 1 - Equilateral Triangles

Let us consider two equilateral triangles, $\triangle PQR$ and $\triangle ABC$.
In any equilateral triangle, each interior angle is exactly $60^\circ$.
Since all angles in both triangles are $60^\circ$, the condition for equal corresponding angles is satisfied.
Furthermore, the ratio of corresponding sides $\frac{PQ}{AB} = \frac{QR}{BC} = \frac{RP}{CA}$ is constant.
Therefore, any two equilateral triangles are similar figures.

Step 2: Identifying Example 2 - Circles

Let us consider two circles with radii $r_1$ and $r_2$ respectively.
A circle is defined by the set of all points in a plane that are at a fixed distance (radius) from a fixed point (center).
Because all circles have the same round shape and lack corners, they are geometrically similar.
The ratio of their circumferences ($2\pi r_1 : 2\pi r_2$) and the ratio of their diameters ($2r_1 : 2r_2$) both simplify to the ratio of their radii ($r_1 : r_2$).
Therefore, any two circles are similar figures.

Final Answer: Two examples of pairs of similar figures are:
1. Any two equilateral triangles.
2. Any two circles.

Given: Two triangles, $\triangle MNL$ and $\triangle QPR$.

In $\triangle MNL$: $MN = 2.5$, $ML = 5$, and $\angle M = 70^\circ$.

In $\triangle QPR$: $PQ = 5$, $QR = 10$, and $\angle Q = 70^\circ$.

To Find: Determine if the triangles are similar, state the similarity criterion, and write the symbolic form.

Step 1: Analyze the ratios of the sides including the given angle.

In $\triangle MNL$ and $\triangle QPR$, the sides forming the angle $70^\circ$ are $(MN, ML)$ and $(PQ, QR)$ respectively.

Calculate the ratio of the corresponding sides:

Ratio 1: $\frac{MN}{PQ} = \frac{2.5}{5} = \frac{1}{2}$

Ratio 2: $\frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}$

[Since $\frac{2.5}{5} = 0.5$ and $\frac{5}{10} = 0.5$]

Step 2: Compare the included angles.

We are given that $\angle M = 70^\circ$ and $\angle Q = 70^\circ$.

Therefore, $\angle M = \angle Q = 70^\circ$.

Step 3: Apply the SAS (Side-Angle-Side) Similarity Criterion.

The SAS Similarity Criterion states that if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Since $\frac{MN}{PQ} = \frac{ML}{QR}$ and $\angle M = \angle Q$, the triangles are similar.

Step 4: Write the symbolic form.

The correspondence is $M \leftrightarrow Q$, $N \leftrightarrow P$, and $L \leftrightarrow R$.

Thus, $\triangle MNL \sim \triangle QPR$.

Final Answer: The triangles are similar by the SAS similarity criterion. The symbolic form is $\triangle MNL \sim \triangle QPR$.

Given: A triangle $PQR$ where $E$ is a point on side $PQ$ and $F$ is a point on side $PR$. The lengths are provided as follows:

• $PE = 3.9 \text{ cm}$
• $EQ = 3 \text{ cm}$
• $PF = 3.6 \text{ cm}$
• $FR = 2.4 \text{ cm}$

To Find: Determine whether $EF \parallel QR$.

Theoretical Basis: According to the Converse of Thales' Theorem (Basic Proportionality Theorem), if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. That is, $EF \parallel QR$ if and only if:

$$\frac{PE}{EQ} = \frac{PF}{FR}$$

Step 1: Calculate the ratio of the segments on side $PQ$.

Given $PE = 3.9 \text{ cm}$ and $EQ = 3 \text{ cm}$.

$$\frac{PE}{EQ} = \frac{3.9}{3}$$

$$\frac{PE}{EQ} = 1.3$$

Step 2: Calculate the ratio of the segments on side $PR$.

Given $PF = 3.6 \text{ cm}$ and $FR = 2.4 \text{ cm}$.

$$\frac{PF}{FR} = \frac{3.6}{2.4}$$

To simplify, multiply both numerator and denominator by 10:

$$\frac{PF}{FR} = \frac{36}{24}$$

Divide both by their greatest common divisor, which is 12:

$$\frac{PF}{FR} = \frac{3}{2} = 1.5$$

Step 3: Compare the ratios.

From Step 1, $\frac{PE}{EQ} = 1.3$.

From Step 2, $\frac{PF}{FR} = 1.5$.

Since $1.3 \neq 1.5$, it follows that:

$$\frac{PE}{EQ} \neq \frac{PF}{FR}$$

Conclusion: Since the ratios of the segments are not equal, the condition for the Converse of the Basic Proportionality Theorem is not satisfied.

Final Answer: No, $EF$ is not parallel to $QR$.

Given:

1. $\triangle ABC \sim \triangle PQR$

2. $AD$ is the median of $\triangle ABC$ to side $BC$ (i.e., $BD = DC = \frac{1}{2}BC$)

3. $PM$ is the median of $\triangle PQR$ to side $QR$ (i.e., $QM = MR = \frac{1}{2}QR$)

To Prove:

$\frac{AB}{PQ} = \frac{AD}{PM}$

Step 1: Utilizing the property of similar triangles

Since $\triangle ABC \sim \triangle PQR$, by the definition of similar triangles, their corresponding sides are proportional and their corresponding angles are equal:

$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ --- (Equation 1)

$\angle B = \angle Q$ --- (Equation 2)

Step 2: Expressing sides in terms of medians

Given that $AD$ and $PM$ are medians, $D$ is the midpoint of $BC$ and $M$ is the midpoint of $QR$.

Therefore, $BC = 2BD$ and $QR = 2QM$.

Substituting these into Equation 1:

$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM}$

Thus, $\frac{AB}{PQ} = \frac{BD}{QM}$ --- (Equation 3)

Step 3: Proving similarity of $\triangle ABD$ and $\triangle PQM$

In $\triangle ABD$ and $\triangle PQM$:

1. $\frac{AB}{PQ} = \frac{BD}{QM}$ [From Equation 3]

2. $\angle B = \angle Q$ [From Equation 2]

By the SAS (Side-Angle-Side) Similarity Criterion, if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the two triangles are similar.

Therefore, $\triangle ABD \sim \triangle PQM$.

Step 4: Establishing the final ratio

Since $\triangle ABD \sim \triangle PQM$, the ratios of their corresponding sides must be equal:

$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$

From this, we extract the required equality:

$\frac{AB}{PQ} = \frac{AD}{PM}$

Final Answer:
Hence, it is proved that $\frac{AB}{PQ} = \frac{AD}{PM}$.

Given: Two triangles, $\triangle ABC$ and $\triangle DEF$.

In $\triangle ABC$:

• $AB = 2.5$
• $BC = 3$
• $\angle A = 80^\circ$

In $\triangle DEF$:

• $EF = 6$
• $DF = 5$
• $\angle F = 80^\circ$

To Find: Determine if the triangles are similar, state the criterion used, and write the symbolic representation if they are similar.

Step 1: Analyze the sides and angles of the triangles.

We examine the ratios of the sides that include the given angle. In $\triangle ABC$, the angle $80^\circ$ is at vertex $A$. The sides forming this angle are $AB$ and $AC$. However, the length of $AC$ is not provided. In $\triangle DEF$, the angle $80^\circ$ is at vertex $F$. The sides forming this angle are $EF$ and $DF$.

Step 2: Check for similarity criteria.

For two triangles to be similar by the SAS (Side-Angle-Side) criterion, the ratio of the two sides must be equal, and the included angle must be equal.

Let us check the ratios of the given sides:

Ratio 1: $\frac{AB}{DF} = \frac{2.5}{5} = 0.5$

Ratio 2: $\frac{BC}{EF} = \frac{3}{6} = 0.5$

Here, the ratios of the sides are equal ($\frac{AB}{DF} = \frac{BC}{EF} = \frac{1}{2}$).

Step 3: Evaluate the included angle condition.

For SAS similarity, the equal angle must be the included angle between the proportional sides.

In $\triangle ABC$, the sides are $AB$ and $BC$. The included angle is $\angle B$. We are given $\angle A = 80^\circ$.

In $\triangle DEF$, the sides are $DF$ and $EF$. The included angle is $\angle F$. We are given $\angle F = 80^\circ$.

Since the given angle $\angle A$ in $\triangle ABC$ is not the included angle between sides $AB$ and $BC$, and we do not have information about $\angle B$ or $\angle F$ being the included angle for the proportional sides in both triangles, the SAS criterion cannot be applied.

Step 4: Conclusion.

Because the equal angles ($80^\circ$) are not the included angles between the sides that are in the same ratio, the triangles do not satisfy the SAS similarity criterion. There is no other information provided to satisfy AA or SSS criteria.

Final Answer: The triangles are not similar.

Given:

In $\triangle ABC$, $AD$ is the altitude to side $BC$ (i.e., $AD \perp BC$) and $CE$ is the altitude to side $AB$ (i.e., $CE \perp AB$). The altitudes $AD$ and $CE$ intersect at point $P$.

To Prove:

$\triangle AEP \sim \triangle ADB$

Step 1: Identify the triangles to be compared.

We are considering $\triangle AEP$ and $\triangle ADB$.

Step 2: List the corresponding angles and properties.

In $\triangle AEP$ and $\triangle ADB$:

1. $\angle AEP = \angle ADB$

Justification: Since $CE \perp AB$, $\angle AEP = 90^\circ$. Since $AD \perp BC$, $\angle ADB = 90^\circ$. Thus, both angles are equal to $90^\circ$.

2. $\angle PAE = \angle DAB$

Justification: Both angles refer to the same angle $\angle A$ of the original triangle $\triangle ABC$. This is the common angle shared by both triangles.

Step 3: Apply the Similarity Criterion.

According to the AA (Angle-Angle) Similarity Criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Since we have established:

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

$\angle PAE = \angle DAB$ (Common angle)

Therefore, by the AA similarity criterion:

$\triangle AEP \sim \triangle ADB$

Conclusion:

We have successfully demonstrated that the two triangles satisfy the conditions for similarity based on the equality of their corresponding angles.

Final Answer: Since $\angle AEP = \angle ADB = 90^\circ$ and $\angle PAE = \angle DAB$ (common angle), by AA similarity criterion, $\triangle AEP \sim \triangle ADB$.

Given:

In $\triangle ABC$, we are given that $DE \parallel BC$.

The lengths provided for figure (ii) are:

• $AE = 1.8 \text{ cm}$
• $BD = 7.2 \text{ cm}$
• $EC = 5.4 \text{ cm}$

To Find:

The length of $AD$.

Visual Representation:

Step 1: Stating the Relevant Theorem

According to the Basic Proportionality Theorem (Thales Theorem), if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

Therefore, in $\triangle ABC$, since $DE \parallel BC$, we have:

$\frac{AD}{DB} = \frac{AE}{EC}$

Step 2: Substituting the Given Values

Let $AD = x$. Substituting the known values into the equation:

$\frac{x}{7.2} = \frac{1.8}{5.4}$

Step 3: Solving for $x$

First, simplify the fraction on the right side:

$\frac{1.8}{5.4} = \frac{18}{54} = \frac{1}{3}$

[Since $18 \times 3 = 54$]

Now, the equation becomes:

$\frac{x}{7.2} = \frac{1}{3}$

To isolate $x$, multiply both sides by $7.2$:

$x = \frac{7.2}{3}$

$x = 2.4$

Step 4: Conclusion

Since $x$ represents the length of $AD$, we conclude that $AD = 2.4 \text{ cm}$.

Final Answer: $AD = 2.4 \text{ cm}$

Given: A statement regarding the similarity of specific types of triangles: "All ______ triangles are similar."

To Find: The correct word from the options (isosceles, equilateral) that completes the statement to make it mathematically true.

Step 1: Understanding the Definition of Similar Triangles
Two triangles are said to be similar if:
1. Their corresponding angles are equal.
2. Their corresponding sides are in the same ratio (proportion).

Step 2: Analyzing Equilateral Triangles
An equilateral triangle is a triangle in which all three sides are equal in length and all three interior angles are equal to $60^\circ$.
Let $T_1$ and $T_2$ be two equilateral triangles with side lengths $s_1$ and $s_2$ respectively.
- The angles of $T_1$ are $60^\circ, 60^\circ, 60^\circ$.
- The angles of $T_2$ are $60^\circ, 60^\circ, 60^\circ$.
Since the corresponding angles are equal ($60^\circ = 60^\circ$), the condition for similarity is satisfied regardless of the side lengths. Thus, all equilateral triangles are similar.

Step 3: Analyzing Isosceles Triangles
An isosceles triangle is a triangle with at least two equal sides. Consider two isosceles triangles:
- Triangle 1: Sides $5, 5, 8$. The angles are approximately $36.87^\circ, 71.56^\circ, 71.56^\circ$.
- Triangle 2: Sides $5, 5, 2$. The angles are approximately $151.04^\circ, 14.48^\circ, 14.48^\circ$.
Since the corresponding angles are not equal, isosceles triangles are not necessarily similar.

Step 4: Conclusion
Based on the geometric properties established in Step 2, the condition of similarity holds true for all equilateral triangles because their angular configuration is constant ($60^\circ$ each).

Final Answer: All equilateral triangles are similar.

Given:

In $\triangle ABC$ and $\triangle PQR$, $AD$ is the median to side $BC$ and $PM$ is the median to side $QR$.

The sides and medians are proportional such that: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.

To Prove:

$\triangle ABC \sim \triangle PQR$

Step 1: Analyzing the given ratios

We are given: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$

Since $AD$ and $PM$ are medians, $D$ is the midpoint of $BC$ and $M$ is the midpoint of $QR$.

Therefore, $BC = 2BD$ and $QR = 2QM$.

Substituting these into the ratio: $\frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM}$.

Thus, the given condition becomes: $\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.

Step 2: Proving $\triangle ABD \sim \triangle PQM$

In $\triangle ABD$ and $\triangle PQM$:

$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$ [Derived in Step 1]

By the SSS (Side-Side-Side) similarity criterion, $\triangle ABD \sim \triangle PQM$.

Consequently, the corresponding angles are equal: $\angle B = \angle Q$ [Since corresponding parts of similar triangles are equal (CPST)].

Step 3: Proving $\triangle ABC \sim \triangle PQR$

In $\triangle ABC$ and $\triangle PQR$:

1. $\frac{AB}{PQ} = \frac{BC}{QR}$ [Given]

2. $\angle B = \angle Q$ [Proved in Step 2]

By the SAS (Side-Angle-Side) similarity criterion, $\triangle ABC \sim \triangle PQR$.

Final Answer:

Since the ratio of two sides is proportional and the included angle is equal, $\triangle ABC \sim \triangle PQR$ is proved.

Given: In $\triangle ABC$, $DE \parallel BC$. The lengths of the segments are provided as follows: $AD = 1.5\text{ cm}$, $DB = 3\text{ cm}$, and $AE = 1\text{ cm}$.

To find: The length of segment $EC$.

Step 1: Identifying the Applicable Theorem
Since $DE \parallel BC$ in $\triangle ABC$, we apply the Basic Proportionality Theorem (Thales Theorem). The theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

Mathematically, this is expressed as: $$\frac{AD}{DB} = \frac{AE}{EC}$$ [By Basic Proportionality Theorem]

Step 2: Substituting the Given Values
We substitute the known values into the equation: $$\frac{1.5}{3} = \frac{1}{EC}$$

Step 3: Solving for $EC$
To isolate $EC$, we perform cross-multiplication: $$1.5 \times EC = 3 \times 1$$ $$1.5 \times EC = 3$$

Now, divide both sides by $1.5$: $$EC = \frac{3}{1.5}$$

To simplify the division, multiply the numerator and denominator by 10: $$EC = \frac{30}{15}$$ $$EC = 2$$

Step 4: Conclusion
The length of segment $EC$ is calculated to be $2\text{ cm}$.

Final Answer: $EC = 2\text{ cm}$

Given: Two polygons having the same number of sides ($n$).

To Find: The conditions under which these two polygons are considered similar.

Step 1: Understanding the Definition of Similar Polygons

In geometry, two polygons with the same number of sides are defined as similar if and only if they satisfy two specific criteria simultaneously. Similarity implies that the shapes are identical in form but not necessarily in size (i.e., one is a scaled version of the other).

Step 2: Analyzing Condition (a) - Corresponding Angles

For two polygons to be similar, their internal structure must maintain the same "shape." This is preserved if the angles at each corresponding vertex are identical. If the angles were different, the polygons would be distorted relative to each other. Therefore, the corresponding angles must be equal.

Step 3: Analyzing Condition (b) - Corresponding Sides

While the angles define the shape, the sides define the scale. For the polygons to be similar, the ratio of the lengths of any two corresponding sides must be constant. This constant ratio is known as the scale factor. Therefore, the corresponding sides must be proportional.

Step 4: Synthesizing the Statement

Based on the geometric definition of similarity for polygons:

(a) Their corresponding angles are equal.

(b) Their corresponding sides are proportional.

Final Answer: Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional.

Given: A triangle $ABC$ and a point $D$ on the side $BC$ such that $\angle ADC = \angle BAC$.

To Prove: $CA^2 = CB \cdot CD$.

Step 1: Identifying the Triangles to be Compared
To prove the relationship $CA^2 = CB \cdot CD$, we observe that this can be rewritten as the ratio $\frac{CA}{CB} = \frac{CD}{CA}$. This suggests that we should consider the two triangles $\triangle ADC$ and $\triangle BAC$.

Step 2: Establishing Similarity Criteria
We compare $\triangle ADC$ and $\triangle BAC$ based on the following observations:

1. In $\triangle ADC$ and $\triangle BAC$, we are given that $\angle ADC = \angle BAC$. [Given in the problem statement]

2. In $\triangle ADC$ and $\triangle BAC$, the angle $\angle C$ is common to both triangles. That is, $\angle ACD = \angle BCA$. [Common angle]

Step 3: Applying the AA (Angle-Angle) Similarity Criterion
Since two angles of $\triangle ADC$ are equal to two corresponding angles of $\triangle BAC$, by the AA similarity criterion, the triangles are similar.

Therefore, $\triangle ADC \sim \triangle BAC$. [By AA Similarity Criterion]

Step 4: Utilizing the Properties of Similar Triangles
When two triangles are similar, the ratios of their corresponding sides are equal. Based on the correspondence established in Step 3:

$\frac{AD}{BA} = \frac{DC}{AC} = \frac{AC}{BC}$

[Since corresponding sides of similar triangles are proportional]

Step 5: Deriving the Final Equation
We take the relevant parts of the proportionality ratio from Step 4:

$\frac{DC}{AC} = \frac{AC}{BC}$

By performing cross-multiplication:

$AC \cdot AC = DC \cdot BC$

$AC^2 = BC \cdot DC$

Since $AC$ is the same as $CA$ and $BC$ is the same as $CB$, we can rewrite this as:

$CA^2 = CB \cdot CD$

Final Answer: Hence, it is proved that $CA^2 = CB \cdot CD$.