Triangles — Notes

Introduction

You are already familiar with triangles and many of their properties from earlier classes. In Class IX, you learned about congruence of triangles.

Congruent figures are figures that have the exact same shape and the exact same size. If you can perfectly overlap two figures, they are congruent.

This chapter introduces similar figures. These are figures that have the same shape but not necessarily the same size.

• Similar figures: Figures with the same shape but not necessarily the same size.
• Congruent figures: Figures with the same shape and the same size.

Think about a photograph and its smaller, wallet-sized version. They have the same shape (both are rectangles with the same proportions), but different sizes. Thus, they are similar. However, a photograph and its original negative are congruent if they are printed at the same size.

The concept of similarity is crucial for indirect measurements, such as calculating the height of tall mountains or the distance to the moon. These are not measured directly with a tape but are found using principles of similarity.

Similarity of Triangles

When we talk about similarity in the context of triangles, two triangles are considered similar if:

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

This means if triangle $ABC$ is similar to triangle $PQR$, denoted as $\mathrm{△}ABC\sim \mathrm{△}PQR$, then:

• $\mathrm{\angle }A=\mathrm{\angle }P$, $\mathrm{\angle }B=\mathrm{\angle }Q$, $\mathrm{\angle }C=\mathrm{\angle }R$ (Corresponding angles are equal).
• $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CA}{RP}$ (Corresponding sides are proportional).

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

Why is similarity important?

Similar triangles allow us to find unknown lengths or angles in geometric figures indirectly. For example, if you know the height of a person and the length of their shadow, you can use similar triangles formed by the person and their shadow, and another object (like a building) and its shadow, to find the height of the building without measuring it directly. This principle is fundamental in surveying, architecture, and trigonometry.

Remember

For triangles to be similar, both conditions (equal corresponding angles AND proportional corresponding sides) must be met.

Exam-smart

In a case study question, you might be given a scenario with two objects casting shadows. You'll be expected to identify that the sun's rays make the angles of elevation equal, and that both objects and the ground form right-angled triangles. You'd then use the proportionality of sides from similarity to find an unknown height or distance.

Similar Figures

Two figures are similar if they have the same shape but not necessarily the same size. This means that all corresponding angles are equal, and all corresponding sides are in the same ratio. Congruent figures are always similar, but similar figures are not always congruent.

Key Concepts:

• Same Shape: Similar figures look alike, like two circles or two squares.
• Different Size: Similar figures can be enlarged or reduced versions of each other.
• Corresponding Angles: Angles in the same relative position in similar figures are equal.
• Corresponding Sides: Sides in the same relative position in similar figures are proportional.
• Scale Factor: The constant ratio between corresponding sides of similar figures.

Conditions for Similarity of Polygons: Two polygons with the same number of sides are similar if:

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

Example: Consider two photographs of the same monument (e.g., Taj Mahal) taken from different distances. They will have the same shape (angles will be the same), but different sizes. The ratio of lengths of corresponding lines in the two photographs will be constant, which is the scale factor.

Example 1

Given: Two quadrilaterals ABCD and PQRS. To find: Whether they are similar. Solution: Looking at the figures, if we can visually confirm that:

1. $\mathrm{\angle }A=\mathrm{\angle }P$, $\mathrm{\angle }B=\mathrm{\angle }Q$, $\mathrm{\angle }C=\mathrm{\angle }R$, $\mathrm{\angle }D=\mathrm{\angle }S$
2. $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{CD}{RS}=\frac{DA}{SP}$ Then, quadrilateral ABCD is similar to quadrilateral PQRS.

Remember

Both conditions (equal corresponding angles AND proportional corresponding sides) must be met for two polygons to be similar. One condition alone is not sufficient.

Exam-smart

A question might provide two figures (e.g., a square and a rectangle, or a rhombus and a square) and ask if they are similar. You would need to check both the angles and the side ratios. For example, a square and a rhombus might have equal sides in ratio but unequal angles, while a square and a rectangle might have equal angles but unequal side ratios.

Similarity of Triangles

Triangles are a special case of polygons. For two triangles to be similar, they must satisfy the same two conditions:

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

If two triangles have equal corresponding angles, they are called equiangular triangles. A key theorem (Thales' Theorem) states that in two equiangular triangles, the ratio of any two corresponding sides is the same.

Example 2

Given: $\mathrm{△}ABC$ and $\mathrm{△}PQR$ To find: If they are similar. Solution: If we are given:

• $\mathrm{\angle }A=\mathrm{\angle }P$, $\mathrm{\angle }B=\mathrm{\angle }Q$, $\mathrm{\angle }C=\mathrm{\angle }R$
• $\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}$ Then, $\mathrm{△}ABC\sim \mathrm{△}PQR$ (The symbol '$\sim$' denotes similarity).

Remember

Similarity of triangles has specific criteria (AAA, AA, SSS, SAS) which we will cover in the next section. This section introduces the definition.

Exam-smart

A source-based question might provide information about the angles and side lengths of two triangles and ask you to determine if they are similar, or which vertex corresponds to which.

Basic Proportionality Theorem

The Basic Proportionality Theorem (BPT), also known as Thales' Theorem, states that:

"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."

In $\mathrm{△}ABC$, if a line $DE$ is drawn such that $DE\parallel BC$, where $D$ is on $AB$ and $E$ is on $AC$, then:

Proof (Conceptual Understanding): The proof relies on the properties of areas of triangles. Triangles with the same height have areas proportional to their bases. Also, triangles on the same base and between the same parallels have equal areas. By relating the areas of $\mathrm{△}ADE$, $\mathrm{△}BDE$, and $\mathrm{△}DEC$, we can establish the ratio of sides.

Formula: If $DE\parallel BC$ in $\mathrm{△}ABC$, then $\frac{AD}{DB}=\frac{AE}{EC}$.

Example 3

Given: In $\mathrm{△}ABC$, $DE\parallel BC$, $AD=2$ cm, $DB=3$ cm, and $AE=4$ cm. To find: The length of $EC$. Solution: Given: $DE\parallel BC$ in $\mathrm{△}ABC$. To find: Length of $EC$. Solution: Using the Basic Proportionality Theorem:

Substitute the given values:

Rearrange to solve for $EC$:

Answer: $EC=6$ cm.

Example 4

Given: In $\mathrm{△}ABC$, $DE\parallel BC$. To find: Prove that $\frac{AB}{AD}=\frac{AC}{AE}$. Solution: Given: $DE\parallel BC$ in $\mathrm{△}ABC$. To find: Prove $\frac{AB}{AD}=\frac{AC}{AE}$. Solution: Since $DE\parallel BC$, by the Basic Proportionality Theorem:

Take the reciprocal of both sides:

Add 1 to both sides:

Combine the terms on each side:

Recognize that $DB+AD=AB$ and $EC+AE=AC$:

This proves the required result.

Remember

The BPT applies only when a line is parallel to one side of a triangle and intersects the other two sides.

Exam-smart

You will often be given a triangle with a line parallel to one side and some side lengths. You'll need to use BPT to find missing side lengths or to prove relationships between segments. Some problems might ask you to prove a variation, like the one in Example 4.

Converse of Basic Proportionality Theorem

The Converse of the Basic Proportionality 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."

In $\mathrm{△}ABC$, if a line $DE$ intersects $AB$ at $D$ and $AC$ at $E$ such that $\frac{AD}{DB}=\frac{AE}{EC}$, then $DE\parallel BC$.

Proof (Conceptual Understanding): This is proved by contradiction. Assume $DE$ is not parallel to $BC$. Then, draw a line $D{E}^{\mathrm{\prime }}$ parallel to $BC$ which intersects $AC$ at $E^{\mathrm{\prime }}$. By BPT, $\frac{AD}{DB}=\frac{A{E}^{\mathrm{\prime }}}{E^{\mathrm{\prime }}C}$. Since we are given $\frac{AD}{DB}=\frac{AE}{EC}$, we have $\frac{A{E}^{\mathrm{\prime }}}{E^{\mathrm{\prime }}C}=\frac{AE}{EC}$. This implies that $E$ and $E^{\mathrm{\prime }}$ must coincide, meaning $DE$ is indeed parallel to $BC$.

Formula: If $\frac{AD}{DB}=\frac{AE}{EC}$ in $\mathrm{△}ABC$, then $DE\parallel BC$.

Example 5

Given: In $\mathrm{△}PQR$, $S$ is a point on $PQ$ and $T$ is a point on $PR$ such that $PS=3$ cm, $SQ=6$ cm, $PT=4$ cm, and $TR=8$ cm. To find: Whether $ST\parallel QR$. Solution: Given: In $\mathrm{△}PQR$, $S$ on $PQ$, $T$ on $PR$. $PS=3$ cm, $SQ=6$ cm, $PT=4$ cm, $TR=8$ cm. To find: Whether $ST\parallel QR$. Solution: Calculate the ratio $\frac{PS}{SQ}$:

Calculate the ratio $\frac{PT}{TR}$:

Since $\frac{PS}{SQ}=\frac{PT}{TR}$, by the Converse of the Basic Proportionality Theorem:

Answer: Yes, $ST\parallel QR$.

Example 6

Given: In $\mathrm{△}ABC$, $D$ and $E$ are points on sides $AB$ and $AC$ respectively such that $BD=3$ cm, $AD=5$ cm, $CE=4.5$ cm, and $AE=6$ cm. To find: Prove that $DE\parallel BC$. Solution: Given: In $\mathrm{△}ABC$, $D$ on $AB$, $E$ on $AC$. $BD=3$ cm, $AD=5$ cm, $CE=4.5$ cm, $AE=6$ cm. To find: Prove $DE\parallel BC$. Solution: Consider the ratio $\frac{AD}{BD}$:

Consider the ratio $\frac{AE}{CE}$:

(Wait, the ratios are not equal. Let me recheck the example and calculation.) Ah, the provided example in the text had a slight mismatch in values for the calculation of AE/CE. Let's use the values as given and see if the question intended to show non-parallelism or if there's a typo. If we assume the problem intended $BD=3,AD=5$ and $CE=4.5,AE=6$, then: Ratio $\frac{AD}{BD}=\frac{5}{3}$. Ratio $\frac{AE}{CE}=\frac{6}{4.5}=\frac{60}{45}=\frac{4}{3}$. Since $\frac{5}{3}\mathrm{\ne }\frac{4}{3}$, the condition for the converse of BPT is not met. Thus, $DE$ is not parallel to $BC$.

Let's assume the problem meant to state values that do satisfy the Converse of BPT, so we can demonstrate the proof. For instance, if $BD=3,AD=5,CE=2.7,AE=4.5$:

In this case, since $\frac{AD}{BD}=\frac{AE}{CE}$, by the Converse of BPT, $DE\parallel BC$.

Let's proceed with the provided text's example calculation for Example 3 in the reference, which states: Given: In Fig. 6.17 (i), $DE\parallel BC$. $AD=1.5$ cm, $BD=1$ cm, $AE=2.25$ cm. To find: EC. Solution: Given: $DE\parallel BC$ in $\mathrm{△}ABC$. $AD=1.5$ cm, $BD=1$ cm, $AE=2.25$ cm. To find: EC. Solution: Using the Basic Proportionality Theorem:

Substitute the given values:

Rearrange to solve for EC:

Answer: $EC=1.5$ cm.

Try this: In $\mathrm{△}PQR$, $S$ is a point on $PQ$ and $T$ is a point on $PR$ such that $PS=5$ cm, $SQ=10$ cm, $PT=7$ cm. Find the length of $TR$ such that $ST\parallel QR$. Answer: $TR=14$ cm.

Remember

The Converse of BPT is crucial for proving lines are parallel when you are given information about side ratios.

Exam-smart

Questions will often give side lengths and ask you to state if a line is parallel to the base, or require you to find a side length that makes the line parallel to the base using the converse. Example 10 in the reference book is a perfect example of using the converse to prove a quadrilateral is a trapezium.

Criteria for Similarity of Triangles

Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio. Symbolically, if $\mathrm{△}ABC\sim \mathrm{△}DEF$, then $\mathrm{\angle }A=\mathrm{\angle }D$, $\mathrm{\angle }B=\mathrm{\angle }E$, $\mathrm{\angle }C=\mathrm{\angle }F$, and $\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$.

The similarity symbol is '~'.

AAA Similarity Criterion

Theorem 6.3: If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio, and hence the two triangles are similar. This is known as the AAA (Angle-Angle-Angle) similarity criterion.

How it works: If you can show that all three pairs of corresponding angles of two triangles are equal, then you automatically know that their corresponding sides are proportional.

Why it matters: This criterion simplifies proving similarity. Instead of checking both angles and sides, you only need to verify the angles.

Example 1: Given: In $\mathrm{△}ABC$ and $\mathrm{△}DEF$, $\mathrm{\angle }A={60}^{\circ }$, $\mathrm{\angle }B={70}^{\circ }$, $\mathrm{\angle }C={50}^{\circ }$ and $\mathrm{\angle }D={60}^{\circ }$, $\mathrm{\angle }E={70}^{\circ }$, $\mathrm{\angle }F={50}^{\circ }$. To find: Whether $\mathrm{△}ABC\sim \mathrm{△}DEF$. Solution:

1. Observe that $\mathrm{\angle }A=\mathrm{\angle }D$ (both ${60}^{\circ }$).
2. Observe that $\mathrm{\angle }B=\mathrm{\angle }E$ (both ${70}^{\circ }$).
3. Observe that $\mathrm{\angle }C=\mathrm{\angle }F$ (both ${50}^{\circ }$).
4. Since all three pairs of corresponding angles are equal, by the AAA similarity criterion, $\mathrm{△}ABC\sim \mathrm{△}DEF$. Answer: $\mathrm{△}ABC\sim \mathrm{△}DEF$.

Try this:

Given: In $\mathrm{△}PQR$ and $\mathrm{△}XYZ$, $\mathrm{\angle }P={80}^{\circ }$, $\mathrm{\angle }Q={50}^{\circ }$, $\mathrm{\angle }R={50}^{\circ }$ and $\mathrm{\angle }X={80}^{\circ }$, $\mathrm{\angle }Y={50}^{\circ }$, $\mathrm{\angle }Z={50}^{\circ }$. To find: Whether $\mathrm{△}PQR\sim \mathrm{△}XYZ$. Answer: Yes, $\mathrm{△}PQR\sim \mathrm{△}XYZ$ by AAA similarity criterion.

AA Similarity Criterion

Remark: If two angles of one triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle, their third angles will also be equal. Therefore, the AAA similarity criterion can be stated as the AA (Angle-Angle) similarity criterion.

How it works: If you can show that any two pairs of corresponding angles of two triangles are equal, the third pair will automatically be equal due to the angle sum property of a triangle. This means the triangles are similar.

Why it matters: This is the most frequently used criterion because it requires checking the fewest conditions (just two angles).

Example 2: Given: In $\mathrm{△}ABC$ and $\mathrm{△}DEF$, $\mathrm{\angle }B={80}^{\circ }$, $\mathrm{\angle }C={40}^{\circ }$ and $\mathrm{\angle }E={80}^{\circ }$, $\mathrm{\angle }F={40}^{\circ }$. To find: Whether $\mathrm{△}ABC\sim \mathrm{△}DEF$. Solution:

1. We are given $\mathrm{\angle }B=\mathrm{\angle }E={80}^{\circ }$.
2. We are given $\mathrm{\angle }C=\mathrm{\angle }F={40}^{\circ }$.
3. Since two pairs of corresponding angles are equal, by the AA similarity criterion, $\mathrm{△}ABC\sim \mathrm{△}DEF$. (The third angles $\mathrm{\angle }A$ and $\mathrm{\angle }D$ will also be equal: ${180}^{\circ }-{80}^{\circ }-{40}^{\circ }={60}^{\circ }$). Answer: $\mathrm{△}ABC\sim \mathrm{△}DEF$.

Example 3: (From Textbook)

Given: In Fig. 6.29, PQ || RS. To find: Prove that $\mathrm{△}POQ\sim \mathrm{△}SOR$. Solution:

1. Given: PQ || RS.
2. Reasoning for angles: Since PQ || RS, we can consider PS and QR as transversals.
3. $\mathrm{\angle }P=\mathrm{\angle }S$ (Alternate interior angles).
4. $\mathrm{\angle }Q=\mathrm{\angle }R$ (Alternate interior angles).
5. Also, $\mathrm{\angle }POQ=\mathrm{\angle }SOR$ (Vertically opposite angles).
6. Conclusion: Since all three pairs of corresponding angles are equal, $\mathrm{△}POQ\sim \mathrm{△}SOR$ by the AAA similarity criterion. Answer: $\mathrm{△}POQ\sim \mathrm{△}SOR$.

Try this:

Given: In $\mathrm{△}XYZ$ and $\mathrm{△}PQR$, $\mathrm{\angle }Y={75}^{\circ }$ and $\mathrm{\angle }Z={35}^{\circ }$, and in $\mathrm{△}PQR$, $\mathrm{\angle }Q={75}^{\circ }$ and $\mathrm{\angle }R={35}^{\circ }$. To find: Whether $\mathrm{△}XYZ\sim \mathrm{△}PQR$. Answer: Yes, $\mathrm{△}XYZ\sim \mathrm{△}PQR$ by AA similarity criterion.

SSS Similarity Criterion

Theorem 6.4: If in two triangles, the sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal, and hence the two triangles are similar. This is known as the SSS (Side-Side-Side) similarity criterion.

How it works: If you can show that the ratios of the lengths of all three pairs of corresponding sides are equal, then you can conclude that the corresponding angles are also equal.

Why it matters: This criterion allows us to prove similarity when we know the side lengths but not necessarily the angles.

Example 4: Given: In $\mathrm{△}ABC$, $AB=3$, $BC=6$, $CA=8$. In $\mathrm{△}DEF$, $DE=4.5$, $EF=9$, $FD=12$. To find: Whether $\mathrm{△}ABC\sim \mathrm{△}DEF$. Solution:

1. Calculate the ratios of corresponding sides:

• $\frac{AB}{DE}=\frac{3}{4.5}=\frac{30}{45}=\frac{2}{3}$
• $\frac{BC}{EF}=\frac{6}{9}=\frac{2}{3}$
• $\frac{CA}{FD}=\frac{8}{12}=\frac{2}{3}$

2. Since $\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=\frac{2}{3}$, the corresponding sides are proportional.
3. By the SSS similarity criterion, $\mathrm{△}ABC\sim \mathrm{△}DEF$. Answer: $\mathrm{△}ABC\sim \mathrm{△}DEF$.

Example 5: (From Textbook)

Observe Fig. 6.30 and then find $\mathrm{\angle }P$. Given: In $\mathrm{△}ABC$ and $\mathrm{△}PQR$, $AB=3.8$, $BC=6$, $CA=3\sqrt{3}$ and $RQ=7.6$, $QP=12$, $PR=6\sqrt{3}$. (Note: The given side lengths in the textbook diagram for $\mathrm{△}ABC$ seem to be $AB=3.8$, $BC=6$, $CA=3\sqrt{3}$ and for $\mathrm{△}PQR$, $PQ=12$, $QR=7.6$, $RP=6\sqrt{3}$. However, the provided solution uses $AB=3.8$, $BC=6$, $CA=3\sqrt{3}$ and $RQ=7.6$, $QP=12$, $PR=6\sqrt{3}$ which implies a specific correspondence. Let's re-examine the ratios based on vertex correspondence $\mathrm{△}ABC\sim \mathrm{△}RQP$.) To find: $\mathrm{\angle }P$. Solution:

1. Let's check the ratios of sides:

• $\frac{AB}{RQ}=\frac{3.8}{7.6}=\frac{1}{2}$
• $\frac{BC}{QP}=\frac{6}{12}=\frac{1}{2}$
• $\frac{CA}{PR}=\frac{3\sqrt{3}}{6\sqrt{3}}=\frac{1}{2}$

2. Since $\frac{AB}{RQ}=\frac{BC}{QP}=\frac{CA}{PR}=\frac{1}{2}$, the corresponding sides are proportional.
3. Therefore, by the SSS similarity criterion, $\mathrm{△}ABC\sim \mathrm{△}RQP$.
4. This implies that corresponding angles are equal: $\mathrm{\angle }A=\mathrm{\angle }R$, $\mathrm{\angle }B=\mathrm{\angle }Q$, and $\mathrm{\angle }C=\mathrm{\angle }P$.
5. Now find $\mathrm{\angle }C$ in $\mathrm{△}ABC$ using the angle sum property: $\mathrm{\angle }C={180}^{\circ }-\mathrm{\angle }A-\mathrm{\angle }B={180}^{\circ }-{80}^{\circ }-{60}^{\circ }={40}^{\circ }$.
6. Since $\mathrm{\angle }C=\mathrm{\angle }P$, we have $\mathrm{\angle }P={40}^{\circ }$. Answer: $\mathrm{\angle }P={40}^{\circ }$.

Try this:

Given: In $\mathrm{△}PQR$, $PQ=5$ cm, $QR=6$ cm, $RP=7$ cm. In $\mathrm{△}XYZ$, $XY=10$ cm, $YZ=12$ cm, $ZX=14$ cm. To find: Whether $\mathrm{△}PQR\sim \mathrm{△}XYZ$. Answer: Yes, $\mathrm{△}PQR\sim \mathrm{△}XYZ$ because $\frac{PQ}{XY}=\frac{QR}{YZ}=\frac{RP}{ZX}=\frac{1}{2}$.

SAS Similarity Criterion

Theorem 6.5: 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. This is known as the SAS (Side-Angle-Side) similarity criterion.

How it works: If you can show that one pair of corresponding angles is equal, AND the ratio of the lengths of the sides forming that angle in one triangle is the same as the ratio of the lengths of the sides forming the corresponding angle in the other triangle, then the triangles are similar.

Why it matters: This criterion is useful when you have information about one angle and two sides.

Example 6: Given: In $\mathrm{△}ABC$, $AB=2$, $\mathrm{\angle }A={50}^{\circ }$, $AC=4$. In $\mathrm{△}DEF$, $DE=3$, $\mathrm{\angle }D={50}^{\circ }$, $DF=6$. To find: Whether $\mathrm{△}ABC\sim \mathrm{△}DEF$. Solution:

1. Observe that $\mathrm{\angle }A=\mathrm{\angle }D={50}^{\circ }$.
2. Check the ratios of the sides including these angles:

• $\frac{AB}{DE}=\frac{2}{3}$
• $\frac{AC}{DF}=\frac{4}{6}=\frac{2}{3}$

3. Since $\frac{AB}{DE}=\frac{AC}{DF}$ and $\mathrm{\angle }A=\mathrm{\angle }D$, by the SAS similarity criterion, $\mathrm{△}ABC\sim \mathrm{△}DEF$. Answer: $\mathrm{△}ABC\sim \mathrm{△}DEF$.

Example 7: (From Textbook)

Given: In Fig. 6.31, $OA\cdot OB=OC\cdot OD$. Show that $\mathrm{\angle }A=\mathrm{\angle }C$ and $\mathrm{\angle }B=\mathrm{\angle }D$. To find: Proof. Solution:

1. Given: $OA\cdot OB=OC\cdot OD$.
2. Rearrange the given equation: $\frac{OA}{OC}=\frac{OD}{OB}$.
3. Consider vertically opposite angles: $\mathrm{\angle }AOD=\mathrm{\angle }COB$.
4. Apply SAS Similarity: We have a pair of equal angles ($\mathrm{\angle }AOD=\mathrm{\angle }COB$) and the sides including these angles are proportional ($\frac{OA}{OC}=\frac{OD}{OB}$).
5. Therefore, by the SAS similarity criterion, $\mathrm{△}AOD\sim \mathrm{△}COB$.
6. Conclusion: Since the triangles are similar, their corresponding angles are equal. So, $\mathrm{\angle }A=\mathrm{\angle }C$ and $\mathrm{\angle }D=\mathrm{\angle }B$. Answer: Proved.

Try this:

Given: In $\mathrm{△}ABC$, $AB=6$, $\mathrm{\angle }B={70}^{\circ }$, $BC=8$. In $\mathrm{△}PQR$, $PQ=9$, $\mathrm{\angle }Q={70}^{\circ }$, $QR=12$. To find: Whether $\mathrm{△}ABC\sim \mathrm{△}PQR$. Answer: Yes, $\mathrm{△}ABC\sim \mathrm{△}PQR$ because $\frac{AB}{PQ}=\frac{6}{9}=\frac{2}{3}$ and $\frac{BC}{QR}=\frac{8}{12}=\frac{2}{3}$, and $\mathrm{\angle }B=\mathrm{\angle }Q={70}^{\circ }$. So, SAS similarity criterion applies.

Remember

For similarity, the order of vertices is crucial. $\mathrm{△}ABC\sim \mathrm{△}DEF$ means $\mathrm{\angle }A=\mathrm{\angle }D$, $\mathrm{\angle }B=\mathrm{\angle }E$, $\mathrm{\angle }C=\mathrm{\angle }F$, and $\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}$. If the correspondence is different, the triangles might not be similar.

Exam-smart

Case/source questions will provide you with a diagram or a description of two triangles. You'll need to identify which similarity criterion (AAA, AA, SSS, SAS) can be applied by checking angle equalities and side ratios. For example, you might be given parallel lines, meaning alternate interior angles are equal (AA), or a common angle and proportional sides (SAS). You might also be given side lengths and asked to check ratios (SSS). The question will ask to "show that" or "prove that" the triangles are similar, or to find an unknown side or angle using similarity.