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

In $\triangle ABC$: $\angle A = 60^\circ$, $\angle B = 80^\circ$, $\angle C = 40^\circ$.

In $\triangle PQR$: $\angle P = 60^\circ$, $\angle Q = 80^\circ$, $\angle R = 40^\circ$.

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

Step 1: Comparing the corresponding angles of the two triangles.

We observe the angles of $\triangle ABC$ and $\triangle PQR$ as follows:

$\angle A = \angle P = 60^\circ$

$\angle B = \angle Q = 80^\circ$

$\angle C = \angle R = 40^\circ$

Step 2: Applying the Similarity Criterion.

[Theorem: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This is known as the Angle-Angle-Angle (AAA) similarity criterion.]

Since all three corresponding angles are equal, the triangles satisfy the AAA similarity criterion.

Step 3: Writing the symbolic form.

When writing the symbolic form, the order of vertices must correspond to the equality of the angles.

Since $\angle A = \angle P$, $\angle B = \angle Q$, and $\angle C = \angle R$, the correspondence is $A \leftrightarrow P$, $B \leftrightarrow Q$, and $C \leftrightarrow R$.

Therefore, $\triangle ABC \sim \triangle PQR$.

Final Answer: The triangles are similar by the AAA similarity criterion. The symbolic form is $\triangle ABC \sim \triangle PQR$.

Given: In $\triangle ABC$ and $\triangle PQR$, $AD$ and $PM$ are medians to sides $BC$ and $QR$ respectively. The sides are proportional such that:

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

To Prove: $\triangle ABC \sim \triangle PQR$

Step 1: Construction

Extend $AD$ to a point $E$ such that $AD = DE$. Join $EC$. Similarly, extend $PM$ to a point $N$ such that $PM = MN$. Join $NR$.

Step 2: Proving Quadrilaterals ABEC and PQNR are Parallelograms

In quadrilateral $ABEC$, diagonals $AE$ and $BC$ bisect each other at $D$ (since $AD=DE$ by construction and $BD=DC$ because $AD$ is a median). [Since diagonals bisect each other, the quadrilateral is a parallelogram]. Therefore, $AB = EC$ and $AC = BE$.

Similarly, in quadrilateral $PQNR$, diagonals $PN$ and $QR$ bisect each other at $M$. Therefore, $PQ = NR$ and $PR = QN$.

Step 3: Establishing Similarity of $\triangle ABE$ and $\triangle PQN$

We are given $\frac{AB}{PQ} = \frac{AC}{PR} = \frac{AD}{PM}$.

Substituting $AB=EC$, $AC=BE$, $PQ=NR$, $PR=QN$, and multiplying $AD$ and $PM$ by $2$:

$\frac{AB}{PQ} = \frac{BE}{QN} = \frac{2AD}{2PM} = \frac{AE}{PN}$

Thus, $\frac{AB}{PQ} = \frac{BE}{QN} = \frac{AE}{PN}$. By SSS similarity criterion, $\triangle ABE \sim \triangle PQN$.

Therefore, $\angle BAE = \angle QPN$ (Corresponding parts of similar triangles). [Equation 1]

Step 4: Establishing Similarity of $\triangle ACE$ and $\triangle PRN$

Similarly, we can show $\triangle ACE \sim \triangle PRN$ using the same logic. Thus, $\angle CAE = \angle RPN$. [Equation 2]

Step 5: Final Conclusion

Adding [Equation 1] and [Equation 2]:

$\angle BAE + \angle CAE = \angle QPN + \angle RPN$

$\angle BAC = \angle QPR$

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

$\frac{AB}{PQ} = \frac{AC}{PR}$ (Given)

$\angle BAC = \angle QPR$ (Proved above)

By SAS similarity criterion, $\triangle ABC \sim \triangle PQR$.

Final Answer: $\triangle ABC \sim \triangle PQR$ is proved.

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 $\triangle ABC$, 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 triangle $\triangle PQR$ where $S$ is a point on side $PR$ and $T$ is a point on side $QR$. It is given that $\angle P = \angle RTS$.

To Prove: $\triangle RPQ \sim \triangle RTS$.

Step 1: Identifying the triangles to be compared.
We are required to prove the similarity between $\triangle RPQ$ and $\triangle RTS$. Let us examine the vertices of these two triangles:

• Triangle 1: $\triangle RPQ$
• Triangle 2: $\triangle RTS$

Step 2: Analyzing the angles of the triangles.
To prove that two triangles are similar, we can use the Angle-Angle (AA) similarity criterion, which states that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Consider $\triangle RPQ$ and $\triangle RTS$:
1. In $\triangle RPQ$, we have $\angle R$ (which is $\angle PRQ$).
2. In $\triangle RTS$, we have $\angle R$ (which is $\angle TRS$).
Since both triangles share the same vertex $R$, we can state: $\angle PRQ = \angle TRS$ [Common angle to both triangles]

Step 3: Utilizing the given information.
It is explicitly given in the problem statement that: $\angle P = \angle RTS$
In the context of our triangles: $\angle RPQ = \angle RTS$ [Given]

Step 4: Applying the AA Similarity Criterion.
We have established two correspondences between the angles of $\triangle RPQ$ and $\triangle RTS$:
1. $\angle R = \angle R$ [Common angle]
2. $\angle P = \angle RTS$ [Given]

By the Angle-Angle (AA) similarity criterion, if two angles of one triangle are equal to two angles of another triangle, the triangles are similar.
Therefore, $\triangle RPQ \sim \triangle RTS$.

Final Answer: Since $\angle R = \angle R$ (common) and $\angle P = \angle RTS$ (given), by the AA similarity criterion, $\triangle RPQ \sim \triangle RTS$.

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 $CA^2 = CB \cdot CD$.

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: Two triangles, $\triangle LMP$ and $\triangle DEF$.

The side lengths of $\triangle LMP$ are:

• $LM = 2.7\text{ cm}$
• $LP = 3\text{ cm}$
• $MP = 2\text{ cm}$

The side lengths of $\triangle DEF$ are:

• $DE = 4\text{ cm}$
• $EF = 6\text{ cm}$
• $DF = 5\text{ cm}$

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

Step 1: Comparing the ratios of corresponding sides.

To check for similarity using the SSS (Side-Side-Side) criterion, we must verify if the ratios of the corresponding sides are equal.

Let us compare the sides in increasing order of length:

Ratio 1: $\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2} = 0.5$

Ratio 2: $\frac{LM}{EF} = \frac{2.7}{6} = 0.45$

Ratio 3: $\frac{LP}{DF} = \frac{3}{5} = 0.6$

Step 2: Evaluating the ratios.

[Since $\frac{MP}{DE} \neq \frac{LM}{EF} \neq \frac{LP}{DF}$]

The ratios of the corresponding sides are not equal. Specifically, $0.5 \neq 0.45 \neq 0.6$.

Step 3: Conclusion based on Similarity Criteria.

For two triangles to be similar by the SSS criterion, the ratios of all three pairs of corresponding sides must be equal. Since the ratios calculated above are not equal, the triangles $\triangle LMP$ and $\triangle DEF$ do not satisfy the conditions for similarity.

Final Answer: The triangles $\triangle LMP$ and $\triangle DEF$ are not similar because the ratios of their corresponding sides are not equal.

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: 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: 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:

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:

1. In $\triangle PQR$, $\angle 1 = \angle 2$ (where $\angle 1 = \angle PQR$ and $\angle 2 = \angle PRQ$).

2. The ratio condition: $\frac{QR}{QS} = \frac{QT}{PR}$.

To Prove:

$\triangle PQS \sim \triangle TQR$

Step 1: Analyze the properties of $\triangle PQR$

In $\triangle PQR$, we are given that $\angle 1 = \angle 2$.

Let $\angle 1 = \angle PQR$ and $\angle 2 = \angle PRQ$.

Since the angles opposite to equal sides are equal, and conversely, if two angles of a triangle are equal, the sides opposite to them are equal [By the Converse of Isosceles Triangle Theorem].

Therefore, $PQ = PR$.

Step 2: Modify the given ratio

The given equation is: $\frac{QR}{QS} = \frac{QT}{PR}$.

Substitute $PR = PQ$ (derived in Step 1) into the equation:

$\frac{QR}{QS} = \frac{QT}{PQ}$

Rearranging the terms to align with the sides of the triangles we intend to prove similar ($\triangle PQS$ and $\triangle TQR$):

$\frac{QS}{QR} = \frac{PQ}{QT}$

Step 3: Establish Similarity using SAS Criterion

Consider $\triangle PQS$ and $\triangle TQR$:

1. From Step 2, we have $\frac{QS}{QR} = \frac{PQ}{QT}$ (or equivalently $\frac{PQ}{QS} = \frac{QT}{QR}$).

2. Observe the common angle between these sides: $\angle PQS = \angle TQR$ (This is the same angle, $\angle 1$).

Since one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, the triangles are similar [By SAS (Side-Angle-Side) Similarity Criterion].

Conclusion:

Therefore, $\triangle PQS \sim \triangle TQR$.

Final Answer: Since the ratio of the sides including the common angle $\angle Q$ are proportional ($\frac{PQ}{QT} = \frac{QS}{QR}$), $\triangle PQS \sim \triangle TQR$ by the SAS similarity criterion.

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:

• $\triangle ODC \sim \triangle OBA$
• $\angle BOC = 125^{\circ}$
• $\angle CDO = 70^{\circ}$
• $DB$ and $AC$ are straight lines intersecting at $O$.

To Find:

• $\angle DOC$
• $\angle DCO$
• $\angle OAB$

Visual Representation:

Step 1: Finding $\angle DOC$

Since $DB$ is a straight line, the angles $\angle DOC$ and $\angle BOC$ form a linear pair. [Axiom: The sum of angles on a straight line is $180^{\circ}$].

$\angle DOC + \angle BOC = 180^{\circ}$

$\angle DOC + 125^{\circ} = 180^{\circ}$

$\angle DOC = 180^{\circ} - 125^{\circ}$

$\angle DOC = 55^{\circ}$

Step 2: Finding $\angle DCO$

Consider $\triangle ODC$. The sum of the interior angles of a triangle is always $180^{\circ}$. [Theorem: Angle Sum Property of a Triangle].

$\angle ODC + \angle DOC + \angle DCO = 180^{\circ}$

Given $\angle ODC = 70^{\circ}$ and we found $\angle DOC = 55^{\circ}$.

$70^{\circ} + 55^{\circ} + \angle DCO = 180^{\circ}$

$125^{\circ} + \angle DCO = 180^{\circ}$

$\angle DCO = 180^{\circ} - 125^{\circ}$

$\angle DCO = 55^{\circ}$

Step 3: Finding $\angle OAB$

It is given that $\triangle ODC \sim \triangle OBA$. [Definition: If two triangles are similar, their corresponding angles are equal].

Therefore, $\angle OAB = \angle OCD$ (which is the same as $\angle DCO$).

Since $\angle DCO = 55^{\circ}$, it follows that:

$\angle OAB = 55^{\circ}$

Final Answer:

$\angle DOC = 55^{\circ}$, $\angle DCO = 55^{\circ}$, and $\angle OAB = 55^{\circ}$.

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:

$\triangle ABD \sim \triangle CBE$

Step 1: Identifying the triangles to be compared

We are considering $\triangle ABD$ and $\triangle CBE$.

Step 2: Analyzing the angles of the triangles

In $\triangle ABD$ and $\triangle CBE$:

1. Consider $\angle ADB$ and $\angle CEB$:

Since $AD \perp BC$, $\angle ADB = 90^\circ$.

Since $CE \perp AB$, $\angle CEB = 90^\circ$.

Therefore, $\angle ADB = \angle CEB = 90^\circ$.

2. Consider $\angle ABD$ and $\angle CBE$:

Observe that $\angle ABD$ is the same angle as $\angle CBE$ because both represent the angle at vertex $B$ of the original triangle $\triangle ABC$.

Therefore, $\angle ABD = \angle CBE$ (Common angle).

Step 3: Applying the Similarity Criterion

We have established two conditions:

i) $\angle ADB = \angle CEB$ ($90^\circ$ each)

ii) $\angle ABD = \angle CBE$ (Common angle)

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.

Step 4: Conclusion

Since two corresponding angles are equal, we conclude that:

$\triangle ABD \sim \triangle CBE$

Final Answer: Hence, it is proved that $\triangle ABD \sim \triangle CBE$ by the AA similarity criterion.

Given: The requirement to provide two distinct examples of pairs of non-similar figures.

To Find: Two pairs of geometric figures that do not satisfy the criteria for similarity.

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 (i.e., proportional).

If either of these conditions is violated, the figures are classified as non-similar.

Step 1: Analyzing Example 1 (Scalene Triangle and Equilateral Triangle)

Let $T_1$ be a scalene triangle with side lengths $3\text{ cm}, 4\text{ cm},$ and $5\text{ cm}$.
Let $T_2$ be an equilateral triangle with side lengths $4\text{ cm}, 4\text{ cm},$ and $4\text{ cm}$.

Justification:

• The corresponding angles of $T_1$ are not equal to the corresponding angles of $T_2$. In $T_2$, all angles are $60^\circ$, whereas in $T_1$, the angles are approximately $36.87^\circ, 53.13^\circ,$ and $90^\circ$.
• The ratio of corresponding sides is not constant: $\frac{3}{4} \neq \frac{4}{4} \neq \frac{5}{4}$.

Since neither condition for similarity is met, these figures are non-similar.

Step 2: Analyzing Example 2 (Rectangle and Rhombus)

Let $R_1$ be a rectangle with sides $4\text{ cm}$ and $2\text{ cm}$.
Let $R_2$ be a rhombus with all sides equal to $3\text{ cm}$.

Justification:

• In a rectangle, all interior angles are $90^\circ$. In a rhombus (that is not a square), the interior angles are not $90^\circ$.
• Because the corresponding angles are not equal, the figures cannot be similar, regardless of the side ratios.

Summary of Findings:

Final Answer: Two examples of pairs of non-similar figures are (1) a scalene triangle and an equilateral triangle, and (2) a rectangle and a rhombus.

Given: A trapezium $ABCD$ in which $AB \parallel DC$. The diagonals $AC$ and $BD$ intersect each other at point $O$.

To Prove: $\frac{OA}{OC} = \frac{OB}{OD}$

Step 1: Identifying the Triangles
Consider $\triangle AOB$ and $\triangle COD$. We aim to prove that these two triangles are similar using the Angle-Angle (AA) similarity criterion.

Step 2: Establishing Equality of Angles
Since $AB \parallel DC$ and $AC$ acts as a transversal, the alternate interior angles are equal. Therefore:
$\angle OAB = \angle OCD$ [Alternate interior angles are equal when lines are parallel]
Similarly, since $AB \parallel DC$ and $BD$ acts as a transversal:
$\angle OBA = \angle ODC$ [Alternate interior angles are equal when lines are parallel]

Step 3: Considering Vertically Opposite Angles
The diagonals $AC$ and $BD$ intersect at $O$. Thus:
$\angle AOB = \angle COD$ [Vertically opposite angles are equal]

Step 4: Applying the Similarity Criterion
In $\triangle AOB$ and $\triangle COD$:
1. $\angle OAB = \angle OCD$ (Proved in Step 2)
2. $\angle OBA = \angle ODC$ (Proved in Step 2)
3. $\angle AOB = \angle COD$ (Proved in Step 3)
By the $AAA$ (Angle-Angle-Angle) similarity criterion, which simplifies to $AA$ similarity:
$\triangle AOB \sim \triangle COD$

Step 5: Establishing the Ratio of Corresponding Sides
Since the triangles are similar, the ratios of their corresponding sides must be equal:
$\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}$ [Corresponding parts of similar triangles are proportional (CPST)]

Step 6: Conclusion
From the proportionality established in Step 5, we extract the required equality:
$\frac{OA}{OC} = \frac{OB}{OD}$

Final Answer: Hence, it is proved that $\frac{OA}{OC} = \frac{OB}{OD}$.

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 each other at point $P$.

To Prove:

$\triangle AEP \sim \triangle CDP$

Step 1: Identifying the triangles and their properties

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

From the given information, $AD \perp BC$ and $CE \perp AB$.

Therefore, $\angle AEP = 90^\circ$ (since $CE \perp AB$) and $\angle CDP = 90^\circ$ (since $AD \perp BC$).

Step 2: Establishing Equality of Angles

In $\triangle AEP$ and $\triangle CDP$:

1. $\angle AEP = \angle CDP = 90^\circ$ [Given that $AD$ and $CE$ are altitudes].

2. $\angle APE = \angle CPD$ [These are vertically opposite angles, which are always equal].

Step 3: Applying the Similarity Criterion

According to the Angle-Angle (AA) 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 that:

$\angle AEP = \angle CDP$ ($90^\circ$ each)

$\angle APE = \angle CPD$ (Vertically opposite angles)

Therefore, by the AA similarity criterion, we conclude:

$\triangle AEP \sim \triangle CDP$

Conclusion:

The triangles $\triangle AEP$ and $\triangle CDP$ satisfy the conditions for similarity as two of their corresponding angles are equal.

Final Answer: Since $\angle AEP = \angle CDP = 90^\circ$ and $\angle APE = \angle CPD$ (vertically opposite angles), by AA similarity criterion, $\triangle AEP \sim \triangle CDP$.