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: 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:

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:

• $\triangle ODC \sim \triangle OBA$
• $\angle BOC = 125^{\circ}$
• $\angle CDO = 70^{\circ}$
• $DB$ 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: 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, $\triangle AEP \sim \triangle CDP$.

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: $\triangle ABE \cong \triangle ACD$.

To Prove: $\triangle ADE \sim \triangle ABC$.

Visual Representation:

Step 1: Analyze the given congruence.

We are given that $\triangle ABE \cong \triangle ACD$. By the property of Congruent Triangles (CPCT - Corresponding Parts of Congruent Triangles), the corresponding sides and angles are equal.

Therefore, $AB = AC$ --- (Equation 1)

And, $AE = AD$ --- (Equation 2)

Step 2: Formulate ratios from the equations.

From Equation 1, we can write: $\frac{AB}{AC} = 1$

From Equation 2, we can write: $\frac{AD}{AE} = 1$

Equating the two, we get: $\frac{AB}{AC} = \frac{AD}{AE}$

By rearranging the terms to align with the sides of $\triangle ADE$ and $\triangle ABC$, we get:

$\frac{AD}{AB} = \frac{AE}{AC}$ --- (Equation 3)

Step 3: Identify the common angle.

In $\triangle ADE$ and $\triangle ABC$, consider the angle at vertex $A$:

$\angle DAE = \angle BAC$ (Common angle to both triangles)

Step 4: Apply the SAS Similarity Criterion.

The SAS (Side-Angle-Side) 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.

From Equation 3, we have the ratio of the sides: $\frac{AD}{AB} = \frac{AE}{AC}$.

From Step 3, we have the included angle: $\angle DAE = \angle BAC$.

Therefore, by SAS similarity criterion:

$\triangle ADE \sim \triangle ABC$

Final Answer: Since the ratio of the corresponding sides is equal and the included angle is common, $\triangle ADE \sim \triangle ABC$ is proved.

Given: Two right-angled triangles $\triangle ABC$ and $\triangle AMP$.

1. $\angle ABC = 90^\circ$ (Given that $\triangle ABC$ is right-angled at $B$).

2. $\angle AMP = 90^\circ$ (Given that $\triangle AMP$ is right-angled at $M$).

To Prove: $\frac{CA}{PA} = \frac{BC}{MP}$

Step 1: Identifying the triangles to be compared.

To prove the ratio $\frac{CA}{PA} = \frac{BC}{MP}$, we must establish the similarity between $\triangle ABC$ and $\triangle AMP$.

Step 2: Analyzing the angles of $\triangle ABC$ and $\triangle AMP$.

In $\triangle ABC$ and $\triangle AMP$:

1. $\angle ABC = \angle AMP = 90^\circ$ (Given, both are right angles).

2. $\angle BAC = \angle MAP$ (This is a common angle shared by both triangles, as vertex $A$ is common to both).

Step 3: Applying the AA (Angle-Angle) Similarity Criterion.

The AA Similarity Criterion states that if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.

Since $\angle ABC = \angle AMP$ and $\angle BAC = \angle MAP$, we conclude:

$\triangle ABC \sim \triangle AMP$ [By AA Similarity Criterion]

Step 4: Using the property of similar triangles.

When two triangles are similar, their corresponding sides are proportional.

For $\triangle ABC \sim \triangle AMP$, the ratios of the corresponding sides are:

$\frac{AB}{AM} = \frac{BC}{MP} = \frac{CA}{PA}$

Step 5: Extracting the required ratio.

From the proportionality established in Step 4, we specifically select the equality:

$\frac{CA}{PA} = \frac{BC}{MP}$

Final Answer: Since $\triangle ABC \sim \triangle AMP$ by the AA similarity criterion, the ratio of their corresponding sides is equal, hence $\frac{CA}{PA} = \frac{BC}{MP}$ is proved.

Given:

Two triangles, $\triangle ABC$ and $\triangle AMP$.

$\angle ABC = 90^\circ$ (Given that $\triangle ABC$ is right-angled at $B$).

$\angle AMP = 90^\circ$ (Given that $\triangle AMP$ is right-angled at $M$).

To Prove:

$\triangle ABC \sim \triangle AMP$

Visual Representation:

Detailed Steps:

Step 1: Identify the triangles to be compared.

We are considering $\triangle ABC$ and $\triangle AMP$.

Step 2: Compare the angles of the two triangles.

In $\triangle ABC$ and $\triangle AMP$:

(i) $\angle ABC = \angle AMP = 90^\circ$ [Given that both are right-angled triangles at $B$ and $M$ respectively].

(ii) $\angle BAC = \angle MAP$ [This is the common angle shared by both triangles, as vertex $A$ is common to both].

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 that:

1. $\angle ABC = \angle AMP$ ($90^\circ$ each)

2. $\angle BAC = \angle MAP$ (Common angle)

Therefore, by the AA similarity criterion, we conclude that:

$\triangle ABC \sim \triangle AMP$

Final Answer:

Since two corresponding angles are equal, $\triangle ABC \sim \triangle AMP$ is proved by the AA similarity criterion.

Given:

1. $\triangle ABC$ is an isosceles triangle with $AB = AC$.

2. $E$ is a point on the side $CB$ produced.

3. $AD \perp BC$ (which implies $\angle ADB = \angle ADC = 90^\circ$).

4. $EF \perp AC$ (which implies $\angle EFC = 90^\circ$).

To Prove:

$\triangle ABD \sim \triangle ECF$

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

Since $\triangle ABC$ is isosceles with $AB = AC$, the angles opposite to the equal sides are equal [Theorem: Angles opposite to equal sides of a triangle are equal].

Therefore, $\angle ABC = \angle ACB$ (or $\angle ABD = \angle ECF$).

Step 2: Identify the criteria for similarity

To prove $\triangle ABD \sim \triangle ECF$, we shall 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.

Step 3: Compare $\triangle ABD$ and $\triangle ECF$

In $\triangle ABD$ and $\triangle ECF$:

1. $\angle ABD = \angle ECF$ (Proved in Step 1, as $\angle ABC = \angle ACB$).

2. $\angle ADB = \angle EFC$ [Since $AD \perp BC$ and $EF \perp AC$, both are $90^\circ$].

Step 4: Conclusion of Similarity

Since two angles of $\triangle ABD$ are equal to two corresponding angles of $\triangle ECF$, by the AA similarity criterion:

$\triangle ABD \sim \triangle ECF$

Final Answer:

Hence, it is proved that $\triangle ABD \sim \triangle ECF$ by the AA similarity criterion.

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

In $\triangle ABC$: $AB = 2$, $BC = 2.5$, $AC = 3$.

In $\triangle PQR$: $PQ = 4$, $QR = 6$, $PR = 5$.

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

Step 1: Compare the ratios of corresponding sides.

To check for similarity by the SSS (Side-Side-Side) criterion, we must verify if the ratios of the corresponding sides are equal. We arrange the sides of both triangles in increasing order of length:

Sides of $\triangle ABC$: $AB=2$, $AC=3$, $BC=2.5$.

Sides of $\triangle PQR$: $PQ=4$, $PR=5$, $QR=6$.

Step 2: Calculate the ratios.

Ratio 1: $\frac{AB}{QR} = \frac{2}{6} = \frac{1}{3}$

Ratio 2: $\frac{BC}{PR} = \frac{2.5}{5} = \frac{1}{2}$

Ratio 3: $\frac{AC}{PQ} = \frac{3}{4} = 0.75$

Step 3: Analyze the results.

Comparing the ratios: $\frac{1}{3} \neq \frac{1}{2} \neq \frac{3}{4}$.

[Since the ratios of the corresponding sides are not equal, the condition for SSS similarity is not satisfied.]

Step 4: Conclusion.

Because the ratios of the corresponding sides of $\triangle ABC$ and $\triangle PQR$ are not proportional, the triangles are not similar.

Final Answer: The triangles are not similar because the ratios of their corresponding sides are not equal.

Given:

1. $\triangle ABC \sim \triangle FEG$

2. $CD$ is the bisector of $\angle ACB$, such that $D$ lies on $AB$.

3. $GH$ is the bisector of $\angle EGF$, such that $H$ lies on $FE$.

To Prove:

$\triangle DCA \sim \triangle HGF$

Step 1: Analyze the implications of the similarity of triangles.

Since $\triangle ABC \sim \triangle FEG$, by the definition of similar triangles (corresponding angles are equal):

$\angle A = \angle F$

$\angle B = \angle E$

$\angle ACB = \angle FEG$

Step 2: Utilize the angle bisector property.

We are given that $CD$ bisects $\angle ACB$ and $GH$ bisects $\angle FEG$.

Therefore, $\angle ACD = \frac{1}{2} \angle ACB$

And, $\angle FGH = \frac{1}{2} \angle FEG$

Since $\angle ACB = \angle FEG$ [From Step 1], it follows that:

$\frac{1}{2} \angle ACB = \frac{1}{2} \angle FEG$

$\angle ACD = \angle FGH$

Step 3: Establish similarity between $\triangle DCA$ and $\triangle HGF$.

In $\triangle DCA$ and $\triangle HGF$:

1. $\angle A = \angle F$ [Proved in Step 1]

2. $\angle ACD = \angle FGH$ [Proved in Step 2]

Step 4: Conclusion by AA Similarity Criterion.

By 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.

Therefore, $\triangle DCA \sim \triangle HGF$.

Final Answer: Since $\angle A = \angle F$ and $\angle ACD = \angle FGH$, by the AA similarity criterion, $\triangle DCA \sim \triangle HGF$ is proved.