TrianglesClass 10 Mathematics NCERT Solutions

Reason: All circles have the same shape, regardless of their radii. Similarity requires same shape but not necessarily same size. Therefore, all circles are similar.

Reason: All squares have corresponding angles that are equal (all are $90^\circ$). The ratio of their corresponding sides is also always equal. For instance, if two squares have side lengths $a$ and $b$, the ratio of their sides is consistently $a:b$. Thus, all squares are similar.

Reason: Every equilateral triangle has three angles of $60^\circ$. This means any two equilateral triangles have corresponding angles that are equal. The ratio of their corresponding sides will also be constant. Therefore, all equilateral triangles are similar.

Reason: This is the formal definition for the similarity of polygons. Both conditions must be met for two polygons to be considered similar.

Given: Two quadrilaterals, ABCD and PQRS. For quadrilateral PQRS: It is a rhombus with sides $PQ = QR = RS = SP = 1.5 \text{ cm}$. The angles are not right angles. For quadrilateral ABCD: It is a square with sides $AB = BC = CD = DA = 3 \text{ cm}$. All angles are $90^\circ$.

To determine: Whether the two quadrilaterals are similar or not.

Conditions for similarity: For two polygons to be similar, two conditions must be satisfied:

Solution: First, let's check the ratio of the corresponding sides: $\frac{PQ}{AB} = \frac{1.5}{3} = \frac{1}{2}$ $\frac{QR}{BC} = \frac{1.5}{3} = \frac{1}{2}$ $\frac{RS}{CD} = \frac{1.5}{3} = \frac{1}{2}$ $\frac{SP}{DA} = \frac{1.5}{3} = \frac{1}{2}$ The ratio of the corresponding sides is constant, so the sides are proportional.

Next, let's check the corresponding angles: In quadrilateral ABCD, all angles are $90^\circ$. So, $\angle A = \angle B = \angle C = \angle D = 90^\circ$. In quadrilateral PQRS, the angles are clearly not $90^\circ$. For example, $\angle P$ is an acute angle and $\angle Q$ is an obtuse angle.

Since the corresponding angles are not equal (e.g., $\angle P \neq \angle A$), the first condition for similarity is not met.

Final Answer: Since one of the conditions for similarity is not satisfied, the two quadrilaterals are not similar.

Given: In both parts (i) and (ii), we have a triangle ABC with a line DE parallel to the side BC, where D is on side AB and E is on side AC.

Theorem Used: Basic Proportionality Theorem (Thales's 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. So, if $\mathrm{DE} \| \mathrm{BC}$, then $\frac{AD}{DB} = \frac{AE}{EC}$.

(i) Given: In $\triangle ABC$, $AD = 1.5 \text{ cm}$, $DB = 3 \text{ cm}$, $AE = 1 \text{ cm}$.

To Find: EC

Solution: According to the Basic Proportionality Theorem: $\frac{AD}{DB} = \frac{AE}{EC}$ Substituting the given values: $\frac{1.5}{3} = \frac{1}{EC}$ $EC = \frac{3 \times 1}{1.5}$ $EC = \frac{3}{1.5} = 2 \text{ cm}$

Final Answer for (i): $EC = 2 \text{ cm}$.

(ii) Given: In $\triangle ABC$, $DB = 7.2 \text{ cm}$, $AE = 1.8 \text{ cm}$, $EC = 5.4 \text{ cm}$.

To Find: AD

Solution: According to the Basic Proportionality Theorem: $\frac{AD}{DB} = \frac{AE}{EC}$ Substituting the given values: $\frac{AD}{7.2} = \frac{1.8}{5.4}$ $\frac{AD}{7.2} = \frac{18}{54} = \frac{1}{3}$ $AD = \frac{7.2}{3}$ $AD = 2.4 \text{ cm}$

Final Answer for (ii): $AD = 2.4 \text{ cm}$.

Given: E and F are points on the sides PQ and PR respectively of a $\triangle PQR$.

Theorem Used: Converse of 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. We need to check if $\frac{PE}{EQ} = \frac{PF}{FR}$. If this condition is true, then $\mathrm{EF} \| \mathrm{QR}$.

(i) Given: $PE = 3.9 \text{ cm}, EQ = 3 \text{ cm}, PF = 3.6 \text{ cm}, FR = 2.4 \text{ cm}$.

Solution: Calculate the ratios: $\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$ $\frac{PF}{FR} = \frac{3.6}{2.4} = \frac{36}{24} = \frac{3}{2} = 1.5$ Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$ ($1.3 \neq 1.5$), the condition for the converse of BPT is not met.

Final Answer for (i): EF is not parallel to QR.

(ii) Given: $PE = 4 \text{ cm}, QE = 4.5 \text{ cm}, PF = 8 \text{ cm}, RF = 9 \text{ cm}$.

Solution: Calculate the ratios: $\frac{PE}{QE} = \frac{4}{4.5} = \frac{40}{45} = \frac{8}{9}$ $\frac{PF}{RF} = \frac{8}{9}$ Since $\frac{PE}{QE} = \frac{PF}{RF}$, the line segment EF divides sides PQ and PR in the same ratio.

Final Answer for (ii): By the converse of the Basic Proportionality Theorem, EF is parallel to QR.

(iii) Given: $PQ = 1.28 \text{ cm}, PR = 2.56 \text{ cm}, PE = 0.18 \text{ cm}, PF = 0.36 \text{ cm}$.

Solution: First, calculate the lengths of EQ and FR. $EQ = PQ - PE = 1.28 - 0.18 = 1.10 \text{ cm}$ $FR = PR - PF = 2.56 - 0.36 = 2.20 \text{ cm}$ Now, calculate the ratios: $\frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55}$ $\frac{PF}{FR} = \frac{0.36}{2.20} = \frac{36}{220} = \frac{9}{55}$ Since $\frac{PE}{EQ} = \frac{PF}{FR}$, the line segment EF divides sides PQ and PR in the same ratio.

(Alternatively, using a corollary of BPT, we can check $\frac{PE}{PQ} = \frac{PF}{PR}$) $\frac{PE}{PQ} = \frac{0.18}{1.28} = \frac{18}{128} = \frac{9}{64}$ $\frac{PF}{PR} = \frac{0.36}{2.56} = \frac{36}{256} = \frac{9}{64}$ This also shows the ratios are equal.

Final Answer for (iii): By the converse of the Basic Proportionality Theorem, EF is parallel to QR.

Given: In a figure, we have a triangle ABC and a triangle ADC, with a common side AC. M is a point on side AB, N is a point on side AD, and L is a point on the common side AC. It is given that LM is parallel to CB ($\mathrm{LM} \| \mathrm{CB}$) and LN is parallel to CD ($\mathrm{LN} \| \mathrm{CD}$).

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

Proof: First, consider the triangle $\triangle ABC$. We are given that $\mathrm{LM} \| \mathrm{CB}$. By the Basic Proportionality Theorem (BPT), if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio. A corollary of this theorem states that the ratio of the parts of the sides is equal to the ratio of the full sides. Applying this corollary to $\triangle ABC$: $\frac{\mathrm{AM}}{\mathrm{AB}} = \frac{\mathrm{AL}}{\mathrm{AC}} \quad \text{...(1)}$

Next, consider the triangle $\triangle ADC$. We are given that $\mathrm{LN} \| \mathrm{CD}$. Applying the same corollary of the BPT to $\triangle ADC$: $\frac{\mathrm{AN}}{\mathrm{AD}} = \frac{\mathrm{AL}}{\mathrm{AC}} \quad \text{...(2)}$

From equation (1) and equation (2), we can see that both ratios $\frac{\mathrm{AM}}{\mathrm{AB}}$ and $\frac{\mathrm{AN}}{\mathrm{AD}}$ are equal to the same ratio $\frac{\mathrm{AL}}{\mathrm{AC}}$. Therefore, we can equate them: $\frac{\mathrm{AM}}{\mathrm{AB}} = \frac{\mathrm{AN}}{\mathrm{AD}}$

Given: In a triangle ABC, D is a point on side AB and E is a point on side BC. A line segment AE is drawn. F is a point on BE. It is given that DE is parallel to AC ($\mathrm{DE} \| \mathrm{AC}$) and DF is parallel to AE ($\mathrm{DF} \| \mathrm{AE}$).

To Prove: $\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}$

Proof: First, consider the triangle $\triangle ABC$. We are given that $\mathrm{DE} \| \mathrm{AC}$. By the Basic Proportionality Theorem (BPT), since DE divides sides BA and BC, we have: $\frac{\mathrm{BD}}{\mathrm{DA}} = \frac{\mathrm{BE}}{\mathrm{EC}} \quad \text{...(1)}$

Next, consider the triangle $\triangle ABE$. We are given that $\mathrm{DF} \| \mathrm{AE}$. By the Basic Proportionality Theorem (BPT), since DF divides sides BA and BE, we have: $\frac{\mathrm{BD}}{\mathrm{DA}} = \frac{\mathrm{BF}}{\mathrm{FE}} \quad \text{...(2)}$

From equation (1) and equation (2), we observe that both ratios $\frac{\mathrm{BE}}{\mathrm{EC}}$ and $\frac{\mathrm{BF}}{\mathrm{FE}}$ are equal to the same ratio $\frac{\mathrm{BD}}{\mathrm{DA}}$. Therefore, we can equate them: $\frac{\mathrm{BF}}{\mathrm{FE}} = \frac{\mathrm{BE}}{\mathrm{EC}}$

Given: In a triangle PQR, O is a point. Lines OP, OQ, and OR are drawn. D is a point on side PQ, E is a point on side PR. A point F is also considered. The problem statement refers to a figure where DE || OQ and DF || OR. This implies a specific geometric configuration, typically where O is a vertex and PQR is a triangle, with D, E, F on the sides OP, OQ, OR. However, solving with the given text suggests D, E, F are on the sides of $\triangle PQR$. Let's assume the standard interpretation where O is a common vertex for three triangles. Let O be a point, and P, Q, R be three other points such that $\triangle OPQ$, $\triangle OQR$, $\triangle OPR$ are formed. Let E be a point on OQ and F be a point on OR. Let D be a point on OP. Given: $\mathrm{DE} \| \mathrm{PQ}$ and $\mathrm{DF} \| \mathrm{PR}$.

To Prove: $\mathrm{EF} \| \mathrm{QR}$

Proof: First, consider the triangle $\triangle OPQ$. We are given that $\mathrm{DE} \| \mathrm{PQ}$. By the Basic Proportionality Theorem (BPT), the line DE divides the sides OP and OQ in the same ratio. $\frac{\mathrm{OD}}{\mathrm{DP}} = \frac{\mathrm{OE}}{\mathrm{EQ}} \quad \text{...(1)}$

Next, consider the triangle $\triangle OPR$. We are given that $\mathrm{DF} \| \mathrm{PR}$. By the Basic Proportionality Theorem (BPT), the line DF divides the sides OP and OR in the same ratio. $\frac{\mathrm{OD}}{\mathrm{DP}} = \frac{\mathrm{OF}}{\mathrm{FR}} \quad \text{...(2)}$

From equation (1) and equation (2), we can equate the ratios: $\frac{\mathrm{OE}}{\mathrm{EQ}} = \frac{\mathrm{OF}}{\mathrm{FR}}$

Now, consider the triangle $\triangle OQR$. We have shown that the line segment EF divides the sides OQ and OR in the same ratio, i.e., $\frac{\mathrm{OE}}{\mathrm{EQ}} = \frac{\mathrm{OF}}{\mathrm{FR}}$. By the Converse of the 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. Therefore, $\mathrm{EF} \| \mathrm{QR}$.

Given: Let O be a point, and P, Q, R be three other points. Lines OP, OQ, and OR are drawn. A, B, and C are points on OP, OQ, and OR, respectively. It is given that AB is parallel to PQ ($\mathrm{AB} \| \mathrm{PQ}$) and AC is parallel to PR ($\mathrm{AC} \| \mathrm{PR}$).

To Prove: BC is parallel to QR ($\mathrm{BC} \| \mathrm{QR}$).

Proof: First, consider the triangle $\triangle OPQ$. We are given that $\mathrm{AB} \| \mathrm{PQ}$. By the Basic Proportionality Theorem (BPT), since the line AB intersects sides OP and OQ, it divides them in the same ratio. $\frac{\mathrm{OA}}{\mathrm{AP}} = \frac{\mathrm{OB}}{\mathrm{BQ}} \quad \text{...(1)}$

Next, consider the triangle $\triangle OPR$. We are given that $\mathrm{AC} \| \mathrm{PR}$. By the Basic Proportionality Theorem (BPT), since the line AC intersects sides OP and OR, it divides them in the same ratio. $\frac{\mathrm{OA}}{\mathrm{AP}} = \frac{\mathrm{OC}}{\mathrm{CR}} \quad \text{...(2)}$

From equation (1) and equation (2), we can equate the ratios as they are both equal to $\frac{\mathrm{OA}}{\mathrm{AP}}$: $\frac{\mathrm{OB}}{\mathrm{BQ}} = \frac{\mathrm{OC}}{\mathrm{CR}}$

Now, consider the triangle $\triangle OQR$. We have shown that the line segment BC divides the sides OQ and OR in the same ratio, i.e., $\frac{\mathrm{OB}}{\mathrm{BQ}} = \frac{\mathrm{OC}}{\mathrm{CR}}$. By the Converse of the 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. Therefore, $\mathrm{BC} \| \mathrm{QR}$.

Given: A triangle ABC. D is the mid-point of side AB. A line is drawn through D parallel to BC, intersecting side AC at point E. So, $\mathrm{AD} = \mathrm{DB}$ and $\mathrm{DE} \| \mathrm{BC}$.

To Prove: The line DE bisects the third side AC. That is, E is the mid-point of AC, or $\mathrm{AE} = \mathrm{EC}$.

Theorem Used: Theorem 6.1 (Basic Proportionality Theorem - BPT): 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.

Proof: In $\triangle ABC$, we are given that $\mathrm{DE} \| \mathrm{BC}$. According to the Basic Proportionality Theorem (Theorem 6.1), the line DE must divide the sides AB and AC in the same ratio. Therefore, we have: $\frac{\mathrm{AD}}{\mathrm{DB}} = \frac{\mathrm{AE}}{\mathrm{EC}}$

We are also given that D is the mid-point of the side AB. This means that the lengths of the segments AD and DB are equal. $\mathrm{AD} = \mathrm{DB}$

Substituting this into our ratio from the BPT: $\frac{\mathrm{DB}}{\mathrm{DB}} = \frac{\mathrm{AE}}{\mathrm{EC}}$ $1 = \frac{\mathrm{AE}}{\mathrm{EC}}$

Multiplying both sides by EC, we get: $\mathrm{EC} = \mathrm{AE}$

This shows that E is the mid-point of the side AC. Thus, the line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

To Prove: The line joining the mid-points of any two sides of a triangle is parallel to the third side.

Given: A triangle $\triangle ABC$, where D is the mid-point of side AB and E is the mid-point of side AC. A line segment DE joins the points D and E.

Proof: Since D is the mid-point of AB, we have: $AD = DB$ Dividing by DB on both sides, we get: $\frac{AD}{DB} = 1 \quad ...(1)$

Since E is the mid-point of AC, we have: $AE = EC$ Dividing by EC on both sides, we get: $\frac{AE}{EC} = 1 \quad ...(2)$

From equations (1) and (2), we have: $\frac{AD}{DB} = \frac{AE}{EC}$

According to Theorem 6.2 (Converse of the 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.

In $\triangle ABC$, the line DE divides the sides AB and AC in the same ratio. Therefore, the line DE is parallel to the third side BC. $DE \parallel BC$

Given: ABCD is a trapezium in which $\mathrm{AB} \| \mathrm{DC}$. The diagonals AC and BD intersect each other at the point O.

To Show: $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$

Proof: In trapezium ABCD, we are given that $\mathrm{AB} \| \mathrm{DC}$. Now, consider the triangles $\triangle AOB$ and $\triangle COD$.

Since AB is parallel to DC, and AC is a transversal, the alternate interior angles are equal. $\angle OAB = \angle OCD \quad ...(1)$

Similarly, since AB is parallel to DC, and BD is a transversal, the alternate interior angles are equal. $\angle OBA = \angle ODC \quad ...(2)$

The angles at the intersection of the diagonals are vertically opposite angles, so they are equal. $\angle AOB = \angle COD \quad ...(3)$

Using the Angle-Angle-Angle (AAA) similarity criterion, from (1), (2), and (3), we can conclude that $\triangle AOB$ is similar to $\triangle COD$. $\triangle AOB \sim \triangle COD$

Since the triangles are similar, the ratio of their corresponding sides must be equal. $\frac{AO}{CO} = \frac{BO}{DO} = \frac{AB}{CD}$

Taking the first two parts of the equality: $\frac{AO}{CO} = \frac{BO}{DO}$

To get the required form, we can rearrange the terms by cross-multiplication. We can swap the positions of BO and CO: $\frac{AO}{BO} = \frac{CO}{DO}$

Given: A quadrilateral ABCD where the diagonals AC and BD intersect each other at the point O such that $\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}$.

To Show: ABCD is a trapezium.

Proof: The given ratio is: $\frac{AO}{BO} = \frac{CO}{DO}$ This can be rearranged by swapping the positions of BO and CO: $\frac{AO}{CO} = \frac{BO}{DO} \quad ...(1)$

Now, consider the triangles $\triangle AOB$ and $\triangle COD$.

At the intersection of the diagonals, the vertically opposite angles are equal: $\angle AOB = \angle COD \quad ...(2)$

From equation (1), we have the ratio of the sides that include these angles: $\frac{AO}{CO} = \frac{BO}{DO}$

Using the Side-Angle-Side (SAS) similarity criterion, with the ratio of two corresponding sides and the included angle being equal, we can conclude that the triangles are similar. $\triangle AOB \sim \triangle COD$

Since the triangles are similar, their corresponding angles must be equal. $\angle OAB = \angle OCD$ and $\angle OBA = \angle ODC$

Now, let's consider the lines AB and DC and the transversal AC. The angles $\angle OAB$ (which is the same as $\angle CAB$) and $\angle OCD$ (which is the same as $\angle ACD$) are alternate interior angles. Since we have proved that these angles are equal, the lines AB and DC must be parallel. $AB \parallel DC$

A quadrilateral with at least one pair of opposite sides parallel is defined as a trapezium.

Therefore, the quadrilateral ABCD is a trapezium.

This question requires identifying similar triangles from pairs shown in figures. The similarity criterion and the symbolic form of the similarity are to be stated.

(i) Description: In $\triangle ABC$, $\angle A = 60^\circ, \angle B = 80^\circ, \angle C = 40^\circ$. In $\triangle PQR$, $\angle P = 60^\circ, \angle Q = 80^\circ, \angle R = 40^\circ$. Solution: In $\triangle ABC$ and $\triangle PQR$: $\angle A = \angle P = 60^\circ$ $\angle B = \angle Q = 80^\circ$ $\angle C = \angle R = 40^\circ$ Since all corresponding angles are equal, the triangles are similar by the AAA (Angle-Angle-Angle) similarity criterion. Answer: $\triangle ABC \sim \triangle PQR$ by AAA similarity.

(ii) Description: In $\triangle ABC$, sides are $AB=2, BC=2.5, AC=3$. In $\triangle PQR$, sides are $PQ=6, QR=4, RP=5$. Solution: Let's check the ratio of corresponding sides: $\frac{AB}{QR} = \frac{2}{4} = \frac{1}{2}$ $\frac{BC}{RP} = \frac{2.5}{5} = \frac{1}{2}$ $\frac{AC}{PQ} = \frac{3}{6} = \frac{1}{2}$ Since the ratios of all corresponding sides are equal, $\frac{AB}{QR} = \frac{BC}{RP} = \frac{AC}{PQ} = \frac{1}{2}$, the triangles are similar by the SSS (Side-Side-Side) similarity criterion. Answer: $\triangle ABC \sim \triangle QRP$ by SSS similarity.

(iii) Description: In $\triangle LMP$, sides are $LM=2.7, MP=2, LP=3$. In $\triangle DEF$, sides are $DE=4, EF=5, DF=6$. Solution: Let's check the ratio of corresponding sides: $\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2}$ $\frac{LP}{DF} = \frac{3}{6} = \frac{1}{2}$ $\frac{LM}{EF} = \frac{2.7}{5} = \frac{27}{50}$ Since $\frac{MP}{DE} = \frac{LP}{DF} \neq \frac{LM}{EF}$, the ratios of corresponding sides are not equal. Answer: The triangles are not similar.

(iv) Description: In $\triangle MNL$, sides are $MN=2.5, ML=5$ and the included angle is $\angle M = 70^\circ$. In $\triangle PQR$, sides are $PQ=5, QR=10$ and the included angle is $\angle Q = 70^\circ$. Solution: In $\triangle MNL$ and $\triangle PQR$: The included angles are equal: $\angle M = \angle Q = 70^\circ$. Let's check the ratio of the sides including these angles: $\frac{MN}{PQ} = \frac{2.5}{5} = \frac{1}{2}$ $\frac{ML}{QR} = \frac{5}{10} = \frac{1}{2}$ Since the ratios of the sides including the equal angles are equal, $\frac{MN}{PQ} = \frac{ML}{QR}$, the triangles are similar by the SAS (Side-Angle-Side) similarity criterion. Answer: $\triangle MNL \sim \triangle QPR$ by SAS similarity.

(v) Description: In $\triangle ABC$, sides are $AB=2.5, BC=3$ and $\angle A = 80^\circ$. In $\triangle DEF$, sides are $DE=5, DF=6$ and $\angle D = 80^\circ$. Solution: In $\triangle ABC$ and $\triangle DEF$, we have $\angle A = \angle D = 80^\circ$. Let's check the ratio of sides adjacent to this angle. For $\triangle ABC$, the sides are AB and AC. For $\triangle DEF$, the sides are DE and DF. The side AC is not given, instead BC is given. Since BC is not the side adjacent to $\angle A$, the SAS criterion cannot be applied. The given information is not sufficient to conclude similarity. Answer: The triangles are not similar.

(vi) Description: In $\triangle DEF$, $\angle D = 70^\circ, \angle E = 80^\circ$. In $\triangle PQR$, $\angle Q = 80^\circ, \angle R = 30^\circ$. Solution: First, find the third angle in each triangle using the angle sum property (sum of angles in a triangle is $180^\circ$). In $\triangle DEF$: $\angle F = 180^\circ - (\angle D + \angle E) = 180^\circ - (70^\circ + 80^\circ) = 180^\circ - 150^\circ = 30^\circ$. In $\triangle PQR$: $\angle P = 180^\circ - (\angle Q + \angle R) = 180^\circ - (80^\circ + 30^\circ) = 180^\circ - 110^\circ = 70^\circ$. Now compare the angles of $\triangle DEF$ and $\triangle PQR$: $\angle D = \angle P = 70^\circ$ $\angle E = \angle Q = 80^\circ$ $\angle F = \angle R = 30^\circ$ Since all corresponding angles are equal, the triangles are similar by the AAA similarity criterion. Answer: $\triangle DEF \sim \triangle PQR$ by AAA similarity.

Given: Two lines AB and CD intersect at point O. We are given that $\triangle ODC \sim \triangle OBA$, $\angle BOC = 125^\circ$, and $\angle CDO = 70^\circ$.

To Find: $\angle DOC$, $\angle DCO$, and $\angle OAB$.

1. Finding $\angle DOC$: The points D, O, B lie on a straight line. Therefore, $\angle DOC$ and $\angle BOC$ form a linear pair of angles. $\angle DOC + \angle BOC = 180^\circ$ $\angle DOC + 125^\circ = 180^\circ$ $\angle DOC = 180^\circ - 125^\circ$ $\angle DOC = 55^\circ$

2. Finding $\angle DCO$: Now, consider the triangle $\triangle DOC$. The sum of angles in a triangle is $180^\circ$. $\angle DCO + \angle CDO + \angle DOC = 180^\circ$ We know $\angle CDO = 70^\circ$ (given) and we found $\angle DOC = 55^\circ$. $\angle DCO + 70^\circ + 55^\circ = 180^\circ$ $\angle DCO + 125^\circ = 180^\circ$ $\angle DCO = 180^\circ - 125^\circ$ $\angle DCO = 55^\circ$

3. Finding $\angle OAB$: It is given that $\triangle ODC \sim \triangle OBA$. When two triangles are similar, their corresponding angles are equal. The correspondence is D $\leftrightarrow$ B, O $\leftrightarrow$ O, and C $\leftrightarrow$ A. Therefore, $\angle ODC = \angle OBA$, $\angle DOC = \angle BOA$, and $\angle OCD = \angle OAB$.

We need to find $\angle OAB$. From the similarity correspondence, we have: $\angle OAB = \angle OCD$ Since $\angle OCD$ is the same as $\angle DCO$, we have: $\angle OAB = \angle DCO$ From our previous calculation, we know that $\angle DCO = 55^\circ$. Therefore, $\angle OAB = 55^\circ$.

Final Answer: $\angle DOC = 55^\circ$ $\angle DCO = 55^\circ$ $\angle OAB = 55^\circ$

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

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

Proof: In $\triangle \mathrm{AOB}$ and $\triangle \mathrm{COD}$, we have:

Since $\mathrm{AB} \| \mathrm{DC}$, and AC is a transversal, the alternate interior angles are equal. $\angle \mathrm{OAB} = \angle \mathrm{OCD} \quad \text{(Alternate interior angles)}$

Similarly, since $\mathrm{AB} \| \mathrm{DC}$, and BD is a transversal, the alternate interior angles are equal. $\angle \mathrm{OBA} = \angle \mathrm{ODC} \quad \text{(Alternate interior angles)}$

Also, the vertically opposite angles are equal. $\angle \mathrm{AOB} = \angle \mathrm{COD} \quad \text{(Vertically opposite angles)}$

By the AAA (Angle-Angle-Angle) similarity criterion, the two triangles are similar. $\triangle \mathrm{AOB} \sim \triangle \mathrm{COD}$

When two triangles are similar, the ratio of their corresponding sides is equal. $\frac{\mathrm{OA}}{\mathrm{OC}} = \frac{\mathrm{OB}}{\mathrm{OD}} = \frac{\mathrm{AB}}{\mathrm{CD}}$

From this, we can conclude that: $\frac{\mathrm{OA}}{\mathrm{OC}} = \frac{\mathrm{OB}}{\mathrm{OD}}$

Given: In a figure containing $\triangle \mathrm{PQR}$, we have points S on QR and T on the line extending from P to Q. We are given the relation $\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$ and $\angle 1 = \angle 2$. In the context of $\triangle PQR$, $\angle 1$ represents $\angle PQR$ and $\angle 2$ represents $\angle PRQ$.

To Prove: $\triangle \mathrm{PQS} \sim \triangle \mathrm{TQR}$

Proof: First, consider $\triangle \mathrm{PQR}$. It is given that $\angle 1 = \angle 2$, which means $\angle PQR = \angle PRQ$. In a triangle, sides opposite to equal angles are equal. Therefore, in $\triangle \mathrm{PQR}$, we have: $\mathrm{PQ} = \mathrm{PR} \quad ...(1)$

Now, we are given the ratio: $\frac{\mathrm{QR}}{\mathrm{QS}} = \frac{\mathrm{QT}}{\mathrm{PR}} \quad ...(2)$

Substitute the value of PR from equation (1) into equation (2): $\frac{\mathrm{QR}}{\mathrm{QS}} = \frac{\mathrm{QT}}{\mathrm{PQ}}$

To better compare the sides for similarity, let's rearrange this equation: $\mathrm{PQ} \times \mathrm{QR} = \mathrm{QS} \times \mathrm{QT}$ $\frac{\mathrm{PQ}}{\mathrm{QT}} = \frac{\mathrm{QS}}{\mathrm{QR}}$

Now, let's consider the triangles we need to prove similar, $\triangle \mathrm{PQS}$ and $\triangle \mathrm{TQR}$.

Since we have two pairs of corresponding sides in proportion and the included angles are equal, we can use the SAS (Side-Angle-Side) similarity criterion.

By SAS similarity criterion, $\triangle \mathrm{PQS} \sim \triangle \mathrm{TQR}$.

Given: In $\triangle \mathrm{PQR}$, S is a point on side PR and T is a point on side QR such that $\angle \mathrm{P} = \angle \mathrm{RTS}$. This means $\angle \mathrm{QPR} = \angle \mathrm{RTS}$.

To Prove: $\triangle \mathrm{RPQ} \sim \triangle \mathrm{RTS}$

Proof: We need to compare $\triangle \mathrm{RPQ}$ and $\triangle \mathrm{RTS}$.

We have found two pairs of corresponding angles that are equal:

By the AA (Angle-Angle) similarity criterion, if two angles of one triangle are equal to two corresponding angles of another triangle, then the two triangles are similar.

Therefore, $\triangle \mathrm{RPQ} \sim \triangle \mathrm{RTS}$.

Given: In a figure, we have two triangles ABE and ACD such that $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD}$.

To Prove: $\triangle \mathrm{ADE} \sim \triangle \mathrm{ABC}$

Proof: Since it is given that $\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD}$, we can use the property that corresponding parts of congruent triangles are equal (CPCTC).

From the congruence, we have:

Now, let's consider the triangles we need to prove similar: $\triangle \mathrm{ADE}$ and $\triangle \mathrm{ABC}$.

Let's examine the ratio of their sides. From (i) and (ii), we can write: $\mathrm{AD} = \mathrm{AE}$

Divide equation (ii) by equation (i). Since lengths are non-zero, we can do this. $\frac{\mathrm{AD}}{\mathrm{AB}} = \frac{\mathrm{AE}}{\mathrm{AC}}$

Now, let's look at the angle between these sides in both triangles. In $\triangle \mathrm{ADE}$, the angle between sides AD and AE is $\angle \mathrm{DAE}$. In $\triangle \mathrm{ABC}$, the angle between sides AB and AC is $\angle \mathrm{BAC}$.

It is clear from the figure that $\angle \mathrm{DAE}$ and $\angle \mathrm{BAC}$ are the same angle. $\angle \mathrm{DAE} = \angle \mathrm{BAC} \quad \text{(Common angle)}$

So, in $\triangle \mathrm{ADE}$ and $\triangle \mathrm{ABC}$, we have:

By the SAS (Side-Angle-Side) similarity criterion, the two triangles are similar. $\triangle \mathrm{ADE} \sim \triangle \mathrm{ABC}$

Given: In $\triangle \mathrm{ABC}$, AD and CE are altitudes that intersect at point P. This means $\mathrm{AD} \perp \mathrm{BC}$ and $\mathrm{CE} \perp \mathrm{AB}$. Therefore, $\angle \mathrm{ADB} = 90^\circ$ and $\angle \mathrm{CEB} = 90^\circ$. Also, $\angle \mathrm{AEP} = 90^\circ$ and $\angle \mathrm{CDP} = 90^\circ$.

(i) To Prove: $\triangle \mathrm{AEP} \sim \triangle \mathrm{CDP}$

Proof: In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{CDP}$:

(ii) To Prove: $\triangle \mathrm{ABD} \sim \triangle \mathrm{CBE}$

Proof: In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{CBE}$:

(iii) To Prove: $\triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}$

Proof: In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{ADB}$:

(iv) To Prove: $\triangle \mathrm{PDC} \sim \triangle \mathrm{BEC}$

Proof: In $\triangle \mathrm{PDC}$ and $\triangle \mathrm{BEC}$:

Given: A parallelogram ABCD. E is a point on the side AD produced. The line segment BE intersects the side CD at a point F.

To Prove: $\triangle \mathrm{ABE} \sim \triangle \mathrm{CFB}$

Proof: In parallelogram ABCD, opposite sides are parallel. So, AD is parallel to BC. Since AD is produced to E, the entire line segment AE is parallel to BC. $AE \parallel BC$

Now, consider the triangles $\triangle \mathrm{ABE}$ and $\triangle \mathrm{CFB}$.

Now, in $\triangle \mathrm{ABE}$ and $\triangle \mathrm{CFB}$, we have two pairs of corresponding angles that are equal:

By the Angle-Angle (AA) similarity criterion, the two triangles are similar. $\triangle \mathrm{ABE} \sim \triangle \mathrm{CFB}$