Given: In $\triangle ADC$ and $\triangle ABC$, we have $LM \parallel CB$ and $LN \parallel CD$.

To Prove: $\frac{AM}{AB} = \frac{AN}{AD}$

Theorem Used: Basic Proportionality Theorem (Thales Theorem), which 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.

Step 1: Applying BPT in $\triangle ABC$

In $\triangle ABC$, we are given that $LM \parallel CB$.

By the Basic Proportionality Theorem, we have:

$\frac{AM}{AB} = \frac{AL}{AC}$ --- (Equation 1)

[Justification: Since $LM \parallel CB$, the ratio of segments on sides $AB$ and $AC$ are proportional.]

Step 2: Applying BPT in $\triangle ADC$

In $\triangle ADC$, we are given that $LN \parallel CD$.

By the Basic Proportionality Theorem, we have:

$\frac{AN}{AD} = \frac{AL}{AC}$ --- (Equation 2)

[Justification: Since $LN \parallel CD$, the ratio of segments on sides $AD$ and $AC$ are proportional.]

Step 3: Comparing the Equations

From Equation 1, we have $\frac{AM}{AB} = \frac{AL}{AC}$.

From Equation 2, we have $\frac{AN}{AD} = \frac{AL}{AC}$.

Since the right-hand sides of both equations are identical ($\frac{AL}{AC}$), we can equate the left-hand sides.

Therefore, $\frac{AM}{AB} = \frac{AN}{AD}$.

[Justification: Euclid's Axiom - Things which are equal to the same thing are equal to one another.]

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

Given:

In $\triangle PQR$, $E$ is a point on $PQ$ and $F$ is a point on $PR$. The given measurements are:

$PQ = 1.28 \text{ cm}$

$PR = 2.56 \text{ cm}$

$PE = 0.18 \text{ cm}$

$PF = 0.36 \text{ cm}$

To Find:

Determine whether $EF \parallel QR$.

Theorem Used:

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, if $\frac{PE}{EQ} = \frac{PF}{FR}$, then $EF \parallel QR$.

Step 1: Calculate the lengths of segments $EQ$ and $FR$.

Since $E$ lies on $PQ$, $EQ = PQ - PE$.

$EQ = 1.28 \text{ cm} - 0.18 \text{ cm} = 1.10 \text{ cm}$

Since $F$ lies on $PR$, $FR = PR - PF$.

$FR = 2.56 \text{ cm} - 0.36 \text{ cm} = 2.20 \text{ cm}$

Step 2: Calculate the ratios $\frac{PE}{EQ}$ and $\frac{PF}{FR}$.

Ratio 1: $\frac{PE}{EQ} = \frac{0.18}{1.10}$

Multiply numerator and denominator by 100 to simplify: $\frac{18}{110} = \frac{9}{55}$

Ratio 2: $\frac{PF}{FR} = \frac{0.36}{2.20}$

Multiply numerator and denominator by 100 to simplify: $\frac{36}{220}$

Divide both by 4: $\frac{36 \div 4}{220 \div 4} = \frac{9}{55}$

Step 3: Compare the ratios.

Since $\frac{PE}{EQ} = \frac{9}{55}$ and $\frac{PF}{FR} = \frac{9}{55}$, it follows that:

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

Step 4: Conclusion based on the Converse of the Basic Proportionality Theorem.

Because the segments are divided in the same ratio, the line $EF$ must be parallel to the side $QR$ by the Converse of the Basic Proportionality Theorem.

Final Answer: Yes, $EF \parallel QR$.

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

• $PE = 4 \text{ cm}$
• $QE = 4.5 \text{ cm}$
• $PF = 8 \text{ cm}$
• $RF = 9 \text{ cm}$

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

Theorem Used: Converse of Thales Theorem (Basic Proportionality Theorem). The theorem states that 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, if $\frac{PE}{EQ} = \frac{PF}{FR}$, then $EF \parallel QR$.

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

The ratio is given by $\frac{PE}{QE}$.

Substituting the given values:

$\frac{PE}{QE} = \frac{4}{4.5}$

To simplify, multiply the numerator and denominator by $10$:

$\frac{PE}{QE} = \frac{40}{45}$

Divide both by their greatest common divisor, which is $5$:

$\frac{PE}{QE} = \frac{8}{9}$

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

The ratio is given by $\frac{PF}{RF}$.

Substituting the given values:

$\frac{PF}{RF} = \frac{8}{9}$

Step 3: Compare the two ratios.

From Step 1, we have $\frac{PE}{QE} = \frac{8}{9}$.

From Step 2, we have $\frac{PF}{RF} = \frac{8}{9}$.

Since $\frac{PE}{QE} = \frac{PF}{RF}$, the condition for the Converse of the Basic Proportionality Theorem is satisfied.

Conclusion: Because the segments are divided in the same ratio, the line segment $EF$ must be parallel to the side $QR$.

Final Answer: Yes, $EF \parallel QR$ because $\frac{PE}{QE} = \frac{PF}{RF} = \frac{8}{9}$.

Given:

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

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

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

To Prove:

$\frac{CD}{GH} = \frac{AC}{FG}$

Step 1: Utilizing the property of similar triangles.

Since $\triangle ABC \sim \triangle FEG$, their corresponding angles are equal and their corresponding sides are proportional [By definition of similar triangles].

Therefore, $\angle A = \angle F$, $\angle B = \angle E$, and $\angle ACB = \angle FGE$.

Also, $\frac{AC}{FG} = \frac{CB}{GE} = \frac{AB}{FE}$.

Step 2: Analyzing the bisected angles.

Given that $CD$ bisects $\angle ACB$, we have $\angle ACD = \angle BCD = \frac{1}{2} \angle ACB$.

Given that $GH$ bisects $\angle FGE$, we have $\angle FGH = \angle EGH = \frac{1}{2} \angle FGE$.

Since $\angle ACB = \angle FGE$, it follows that $\frac{1}{2} \angle ACB = \frac{1}{2} \angle FGE$.

Thus, $\angle ACD = \angle FGH$.

Step 3: Proving similarity between $\triangle ACD$ and $\triangle FGH$.

Consider $\triangle ACD$ and $\triangle FGH$:

1. $\angle A = \angle F$ [From Step 1, as $\triangle ABC \sim \triangle FEG$]

2. $\angle ACD = \angle FGH$ [From Step 2]

Therefore, $\triangle ACD \sim \triangle FGH$ [By AA similarity criterion].

Step 4: Establishing the ratio of sides.

Since $\triangle ACD \sim \triangle FGH$, the ratio of their corresponding sides must be equal [Property of similar triangles].

$\frac{AC}{FG} = \frac{CD}{GH} = \frac{AD}{FH}$.

Conclusion:

From the equality $\frac{AC}{FG} = \frac{CD}{GH}$, we have successfully proven the required relation.

Final Answer: $\frac{CD}{GH} = \frac{AC}{FG}$

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

In $\triangle DEF$:

• $\angle D = 70^\circ$
• $\angle E = 80^\circ$
• $\angle F = 180^\circ - (70^\circ + 80^\circ) = 30^\circ$ [By Angle Sum Property of a triangle]

In $\triangle PQR$:

• $\angle Q = 80^\circ$
• $\angle R = 30^\circ$
• $\angle P = 180^\circ - (80^\circ + 30^\circ) = 70^\circ$ [By Angle Sum Property of a triangle]

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

Step 1: Comparing the angles of $\triangle DEF$ and $\triangle PQR$

We compare the corresponding angles of the two triangles:

• $\angle D = \angle P = 70^\circ$
• $\angle E = \angle Q = 80^\circ$
• $\angle F = \angle R = 30^\circ$

Step 2: Applying the Similarity Criterion

According to the AAA (Angle-Angle-Angle) Similarity Criterion, 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.

Since all three pairs of corresponding angles are equal, the triangles are similar.

Step 3: Writing the Symbolic Form

The vertices must be written in the order of their correspondence:

Vertex $D$ corresponds to $P$, vertex $E$ corresponds to $Q$, and vertex $F$ corresponds to $R$.

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

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

Given: A set of geometric figures known as squares.

To Find: The correct term to describe the relationship between all squares, choosing between "similar" and "congruent".

Step 1: Defining Similarity of Polygons
Two polygons are said to be similar if:
1. Their corresponding angles are equal.
2. Their corresponding sides are in the same ratio (proportion).
[Definition of Similar Polygons]

Step 2: Analyzing the Properties of Squares
Let $S_1$ and $S_2$ be two squares with side lengths $a_1$ and $a_2$ respectively.
- All interior angles of any square are $90^\circ$. Therefore, the corresponding angles of $S_1$ and $S_2$ are equal ($90^\circ = 90^\circ$).
- The ratio of corresponding sides is $\frac{a_1}{a_2}$. Since all sides of a square are equal, the ratio of all pairs of corresponding sides is constant, i.e., $\frac{a_1}{a_2} = \frac{a_1}{a_2} = \frac{a_1}{a_2} = \frac{a_1}{a_2}$.
[Since all squares satisfy the criteria for similarity]

Step 3: Evaluating Congruence
Two figures are congruent if they have the same shape and the same size. Since squares can have different side lengths (e.g., a square of side 2cm and a square of side 5cm), they are not necessarily congruent.
[Definition of Congruent Figures]

Step 4: Conclusion
Because all squares possess the same shape (equiangular and proportional sides) but not necessarily the same size, they are classified as similar figures.
[Logical deduction based on geometric definitions]

Final Answer: All squares are similar.

Given: In $\triangle ABC$, $AD \perp BC$ and $CE \perp AB$. The altitudes $AD$ and $CE$ intersect each other at point $P$.

To Prove: $\triangle PDC \sim \triangle BEC$.

Visual Representation:

Step 1: Identify the triangles to be compared.
We are considering $\triangle PDC$ and $\triangle BEC$.

Step 2: Analyze the angles of $\triangle PDC$ and $\triangle BEC$.
In $\triangle PDC$ and $\triangle BEC$:

1. Consider $\angle PCD$ and $\angle BCE$.
Since $P$ lies on $CE$ and $D$ lies on $BC$, the angle $\angle PCD$ is the same as $\angle BCE$.
Therefore, $\angle PCD = \angle BCE$ [Common angle].

2. Consider the right angles in both triangles:
In $\triangle PDC$, $AD \perp BC$, therefore $\angle PDC = 90^\circ$.
In $\triangle BEC$, $CE \perp AB$, therefore $\angle BEC = 90^\circ$.
Thus, $\angle PDC = \angle BEC = 90^\circ$ [Given that $AD$ and $CE$ are altitudes].

Step 3: Apply 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.
From Step 2, we have:
$\angle PCD = \angle BCE$ (Common)
$\angle PDC = \angle BEC$ (Each is $90^\circ$)

Step 4: Conclusion of Similarity.
By the AA similarity criterion, since two corresponding angles are equal, the triangles must be similar.
$\triangle PDC \sim \triangle BEC$ [By AA Similarity Criterion].

Final Answer: Since $\angle PCD = \angle BCE$ and $\angle PDC = \angle BEC = 90^\circ$, by the AA similarity criterion, $\triangle PDC \sim \triangle BEC$.