Triangles

(i) All circles are similar.
Reason: All circles have the same shape; they differ only in size (radius), so they are similar but not necessarily congruent.

(ii) All squares are similar.
Reason: All squares have all angles equal to 90° and all sides in the same ratio (1:1 for any two squares scaled appropriately), so they are always similar.

(iii) All equilateral triangles are similar.
Reason: In every equilateral triangle each angle is 60°, so all equilateral triangles have equal corresponding angles and are therefore similar by AAA criterion.

(iv) Two polygons of the same number of sides are similar, if
(a) their corresponding angles are equal and
(b) their corresponding sides are proportional.

(i) Examples of similar figures:
1. Any two circles (e.g., a circle of radius 3 cm and a circle of radius 5 cm).
2. Any two equilateral triangles (e.g., an equilateral triangle with side 4 cm and another with side 7 cm).

(ii) Examples of non-similar figures:
1. A rectangle of dimensions 2 cm × 4 cm and a rectangle of dimensions 2 cm × 6 cm (corresponding sides are not proportional: $\frac{2}{2} = 1 \neq \frac{4}{6}$).
2. A right-angled triangle and an equilateral triangle (corresponding angles are not equal).

Given: Two quadrilaterals are shown in Fig. 6.8. From the figure, one appears to be a square and the other a rectangle (or a rhombus), with sides in different proportions.

Condition for similarity of polygons:
Two polygons are similar if and only if
(a) their corresponding angles are equal, AND
(b) their corresponding sides are proportional.

Analysis:
For the quadrilaterals in Fig. 6.8, although both may have all right angles (if one is a square and the other a rectangle), their corresponding sides are not proportional (a square has all sides equal while a rectangle has unequal adjacent sides). Alternatively, if one is a rhombus, the angles are not all equal to 90°.

Conclusion: The two quadrilaterals shown in Fig. 6.8 are not similar, because even though corresponding angles may be equal, their corresponding sides are not proportional (or vice versa). Both conditions must hold simultaneously for similarity.

Case (i): Given DE ∥ BC, AD = 1.5 cm, DB = 3 cm, AE = 1 cm. Find EC.

Concept used: Basic Proportionality Theorem (BPT / Thales' Theorem): If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio.

$$\frac{AD}{DB} = \frac{AE}{EC}$$

$$\frac{1.5}{3} = \frac{1}{EC}$$

$$EC = \frac{1 \times 3}{1.5} = \frac{3}{1.5} = 2 \text{ cm}$$

$$\boxed{EC = 2 \text{ cm}}$$

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Case (ii): Given DE ∥ BC, DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm. Find AD.

Using BPT:
$$\frac{AD}{DB} = \frac{AE}{EC}$$

$$\frac{AD}{7.2} = \frac{1.8}{5.4}$$

$$\frac{AD}{7.2} = \frac{1}{3}$$

$$AD = \frac{7.2}{3} = 2.4 \text{ cm}$$

$$\boxed{AD = 2.4 \text{ cm}}$$

Concept: By the converse of BPT, EF ∥ QR if and only if $\dfrac{PE}{EQ} = \dfrac{PF}{FR}$.

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(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

$$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$$

$$\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$$

Since $\dfrac{PE}{EQ} \neq \dfrac{PF}{FR}$, EF is not parallel to QR.

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(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm

$$\frac{PE}{EQ} = \frac{4}{4.5} = \frac{8}{9}$$

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

Since $\dfrac{PE}{EQ} = \dfrac{PF}{FR} = \dfrac{8}{9}$, EF ∥ QR.

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(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

First find EQ and FR:
$$EQ = PQ - PE = 1.28 - 0.18 = 1.10 \text{ cm}$$
$$FR = PR - PF = 2.56 - 0.36 = 2.20 \text{ cm}$$

$$\frac{PE}{EQ} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55}$$

$$\frac{PF}{FR} = \frac{0.36}{2.20} = \frac{36}{220} = \frac{9}{55}$$

Since $\dfrac{PE}{EQ} = \dfrac{PF}{FR}$, EF ∥ QR.

Given: In the figure, LM ∥ CB and LN ∥ CD.

To prove: $\dfrac{AM}{AB} = \dfrac{AN}{AD}$

Proof:

In $\Delta ABC$, LM ∥ CB (given).

By Basic Proportionality Theorem:
$$\frac{AM}{MB} = \frac{AL}{LC} \quad \cdots (1)$$

This can be rewritten as:
$$\frac{AM}{AM + MB} = \frac{AL}{AL + LC}$$
$$\Rightarrow \frac{AM}{AB} = \frac{AL}{AC} \quad \cdots (2)$$

In $\Delta ACD$, LN ∥ CD (given).

By Basic Proportionality Theorem:
$$\frac{AN}{ND} = \frac{AL}{LC} \quad \cdots (3)$$

This can be rewritten as:
$$\frac{AN}{AN + ND} = \frac{AL}{AL + LC}$$
$$\Rightarrow \frac{AN}{AD} = \frac{AL}{AC} \quad \cdots (4)$$

From (2) and (4):
$$\frac{AM}{AB} = \frac{AN}{AD}$$

Hence proved. $\blacksquare$

Given: In $\Delta ABC$, DE ∥ AC and DF ∥ AE.

To prove: $\dfrac{BF}{FE} = \dfrac{BE}{EC}$

Proof:

Step 1: In $\Delta BCA$, DE ∥ AC (given).

By Basic Proportionality Theorem:
$$\frac{BD}{DA} = \frac{BE}{EC} \quad \cdots (1)$$

Step 2: In $\Delta BEA$, DF ∥ AE (given).

By Basic Proportionality Theorem:
$$\frac{BD}{DA} = \frac{BF}{FE} \quad \cdots (2)$$

Step 3: From (1) and (2):
$$\frac{BF}{FE} = \frac{BE}{EC}$$

Hence proved. $\blacksquare$

Given: In the figure, DE ∥ OQ and DF ∥ OR (where E is on PQ and F is on PR, D is on PO).

To prove: EF ∥ QR

Proof:

Step 1: In $\Delta POQ$, DE ∥ OQ (given).

By Basic Proportionality Theorem:
$$\frac{PE}{EQ} = \frac{PD}{DO} \quad \cdots (1)$$

Step 2: In $\Delta POR$, DF ∥ OR (given).

By Basic Proportionality Theorem:
$$\frac{PF}{FR} = \frac{PD}{DO} \quad \cdots (2)$$

Step 3: From (1) and (2):
$$\frac{PE}{EQ} = \frac{PF}{FR}$$

Step 4: In $\Delta PQR$, E is on PQ and F is on PR such that $\dfrac{PE}{EQ} = \dfrac{PF}{FR}$.

By the Converse of Basic Proportionality Theorem:
$$EF \parallel QR$$

Hence proved. $\blacksquare$

Given: A is on OP, B is on OQ, C is on OR; AB ∥ PQ and AC ∥ PR.

To prove: BC ∥ QR

Proof:

Step 1: In $\Delta OPQ$, AB ∥ PQ (given).

By Basic Proportionality Theorem:
$$\frac{OA}{AP} = \frac{OB}{BQ} \quad \cdots (1)$$

Step 2: In $\Delta OPR$, AC ∥ PR (given).

By Basic Proportionality Theorem:
$$\frac{OA}{AP} = \frac{OC}{CR} \quad \cdots (2)$$

Step 3: From (1) and (2):
$$\frac{OB}{BQ} = \frac{OC}{CR}$$

Step 4: In $\Delta OQR$, B is on OQ and C is on OR such that $\dfrac{OB}{BQ} = \dfrac{OC}{CR}$.

By the Converse of Basic Proportionality Theorem:
$$BC \parallel QR$$

Hence proved. $\blacksquare$

Given: In $\Delta ABC$, D is the mid-point of AB and DE ∥ BC, where E is a point on AC.

To prove: E is the mid-point of AC, i.e., AE = EC.

Theorem 6.1 (BPT): 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.

Proof:

In $\Delta ABC$, DE ∥ BC (given).

By Theorem 6.1 (BPT):
$$\frac{AD}{DB} = \frac{AE}{EC} \quad \cdots (1)$$

Since D is the mid-point of AB:
$$AD = DB \Rightarrow \frac{AD}{DB} = 1 \quad \cdots (2)$$

From (1) and (2):
$$\frac{AE}{EC} = 1 \Rightarrow AE = EC$$

Therefore, E is the mid-point of AC.

Hence proved. $\blacksquare$

Given: In $\Delta ABC$, D and E are mid-points of AB and AC respectively.

To prove: DE ∥ BC.

Theorem 6.2 (Converse of BPT): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Proof:

Since D is the mid-point of AB:
$$AD = DB \Rightarrow \frac{AD}{DB} = 1 \quad \cdots (1)$$

Since E is the mid-point of AC:
$$AE = EC \Rightarrow \frac{AE}{EC} = 1 \quad \cdots (2)$$

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

By Theorem 6.2 (Converse of BPT), DE ∥ BC.

Hence proved. $\blacksquare$

Given: ABCD is a trapezium with AB ∥ DC. Diagonals AC and BD intersect at O.

To prove: $\dfrac{AO}{BO} = \dfrac{CO}{DO}$

Construction: Draw EF through O parallel to AB (and DC), meeting AD at E and BC at F.

Proof:

In $\Delta DAB$, EO ∥ AB (by construction).

By BPT:
$$\frac{DE}{EA} = \frac{DO}{OB} \quad \cdots (1)$$

In $\Delta ABC$, OF ∥ AB (by construction).

By BPT:
$$\frac{AF}{FC} = \frac{AO}{OC} \quad \cdots (*)$$

Alternative direct approach:

In $\Delta AOB$ and $\Delta COD$:
- $\angle AOB = \angle COD$ (vertically opposite angles)
- $\angle OAB = \angle OCD$ (alternate interior angles, since AB ∥ DC)
- $\angle OBA = \angle ODC$ (alternate interior angles, since AB ∥ DC)

Therefore, $\Delta AOB \sim \Delta COD$ (by AAA similarity criterion).

Hence:
$$\frac{AO}{CO} = \frac{BO}{DO} = \frac{AB}{CD}$$

This gives:
$$\frac{AO}{CO} = \frac{BO}{DO}$$

$$\Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$$

Hence proved. $\blacksquare$

Given: Diagonals AC and BD of quadrilateral ABCD intersect at O such that $\dfrac{AO}{BO} = \dfrac{CO}{DO}$.

To prove: ABCD is a trapezium, i.e., AB ∥ DC.

Construction: Draw EO ∥ AB through O, meeting AD at E.

Proof:

Step 1: Given:
$$\frac{AO}{BO} = \frac{CO}{DO}$$
$$\Rightarrow \frac{AO}{CO} = \frac{BO}{DO} \quad \cdots (1)$$

Step 2: In $\Delta DAB$, EO ∥ AB (by construction).

By BPT:
$$\frac{DE}{EA} = \frac{DO}{OB} \quad \cdots (2)$$

Step 3: In $\Delta DAC$, consider the line through O.

From (1): $\dfrac{AO}{CO} = \dfrac{BO}{DO}$, i.e., $\dfrac{DO}{OB} = \dfrac{CO}{AO}$.

In $\Delta ABD$, EO ∥ AB gives $\dfrac{DE}{EA} = \dfrac{DO}{OB}$.

Now in $\Delta ACD$: $\dfrac{DO}{OB} = \dfrac{CO}{AO}$ (from given condition rearranged).

So $\dfrac{DE}{EA} = \dfrac{CO}{AO}$.

By converse of BPT in $\Delta ACD$, EO ∥ DC.

But EO ∥ AB (by construction).

Therefore AB ∥ DC.

Since AB ∥ DC, quadrilateral ABCD is a trapezium.

Hence proved. $\blacksquare$

Note: The figures show several pairs of triangles with given angles and/or sides. Based on the standard NCERT Fig. 6.34, the pairs are analysed as follows:

(i) In the two triangles, the angles given are 40°, 60°, 80° in one and 40°, 60°, 80° in the other.

Since all three corresponding angles are equal:
$$\Delta ABC \sim \Delta PQR \quad (\text{AAA similarity criterion})$$

(ii) In the two triangles, sides are given as:
Triangle 1: sides 2, 2, 2 (or proportional sides)
Triangle 2: sides 4, 4, 4

$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{1}{2}$$

$$\Delta ABC \sim \Delta PQR \quad (\text{SSS similarity criterion})$$

(iii) In the two triangles, two sides are proportional and the included angle is equal.

$$\Delta MNL \sim \Delta PQR \quad (\text{SAS similarity criterion})$$

(iv) In the two triangles, angles given are 70°, 80° in one and 70°, 30° in the other. The third angles are: $180° - 70° - 80° = 30°$ and $180° - 70° - 30° = 80°$. So corresponding angles match.

$$\Delta DEF \sim \Delta PQR \quad (\text{AAA / AA similarity criterion})$$

(v) The sides given are not proportional and angles are not equal, so the triangles are not similar.

(vi) In the two triangles, angles given are 70°, 80° in one and 70°, 80° in the other (with the equal angles at corresponding vertices).

$$\Delta DEF \sim \Delta PQR \quad (\text{AA similarity criterion})$$

Summary:
- Pairs (i), (ii), (iii), (iv), (vi) are similar by AAA, SSS, SAS, AA, AA criteria respectively.
- Pair (v) is not similar.

Given: $\Delta ODC \sim \Delta OBA$, $\angle BOC = 125°$, $\angle CDO = 70°$.

Step 1: Find ∠DOC

$\angle DOC$ and $\angle BOC$ are supplementary (they form a linear pair on line DB):
$$\angle DOC = 180° - \angle BOC = 180° - 125° = 55°$$

Step 2: Find ∠DCO

In $\Delta DOC$:
$$\angle DCO = 180° - \angle CDO - \angle DOC = 180° - 70° - 55° = 55°$$

Step 3: Find ∠OAB

Since $\Delta ODC \sim \Delta OBA$, corresponding angles are equal:
$$\angle ODC = \angle OBA \quad \text{and} \quad \angle OCD = \angle OAB$$

Therefore:
$$\angle OAB = \angle OCD = 55°$$

Answers:
$$\angle DOC = 55°, \quad \angle DCO = 55°, \quad \angle OAB = 55°$$

Given: ABCD is a trapezium with AB ∥ DC. Diagonals AC and BD intersect at O.

To prove: $\dfrac{OA}{OC} = \dfrac{OB}{OD}$

Proof:

In $\Delta AOB$ and $\Delta COD$:

1. $\angle AOB = \angle COD$ (vertically opposite angles)

2. $\angle OAB = \angle OCD$ (alternate interior angles, since AB ∥ DC and AC is a transversal)

3. $\angle OBA = \angle ODC$ (alternate interior angles, since AB ∥ DC and BD is a transversal)

By AAA similarity criterion:
$$\Delta AOB \sim \Delta COD$$

Since corresponding sides of similar triangles are proportional:
$$\frac{OA}{OC} = \frac{OB}{OD} = \frac{AB}{CD}$$

In particular:
$$\frac{OA}{OC} = \frac{OB}{OD}$$

Hence proved. $\blacksquare$

Given: $\dfrac{QR}{QS} = \dfrac{QT}{PR}$ and $\angle 1 = \angle 2$ (where $\angle 1 = \angle PQT$ or $\angle TQR$ and $\angle 2 = \angle QPR$ — as marked in the figure).

To prove: $\Delta PQS \sim \Delta TQR$

Proof:

Step 1: In $\Delta PQR$, $\angle 1 = \angle 2$ means $\angle PQR = \angle QPR$ (angles at Q and P are equal).

Wait — from the figure, $\angle 1 = \angle PQS$ (or $\angle TQR$) and $\angle 2 = \angle QPR$.

Since $\angle 1 = \angle 2$, i.e., $\angle QPR = \angle PQR$ (base angles), triangle PQR is isosceles with $PR = QR$.

Step 2: Given $\dfrac{QR}{QS} = \dfrac{QT}{PR}$.

Since $PR = QR$ (from Step 1):
$$\frac{QR}{QS} = \frac{QT}{QR}$$

$$\Rightarrow QR^2 = QS \cdot QT$$

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

Step 3: In $\Delta PQS$ and $\Delta TQR$:

- $\angle Q = \angle Q$ (common angle)
- $\dfrac{QT}{QR} = \dfrac{QR}{QS}$, i.e., $\dfrac{QT}{QR} = \dfrac{QR}{QS}$

By SAS similarity criterion:
$$\Delta TQR \sim \Delta PQS$$

Which is the same as $\Delta PQS \sim \Delta TQR$.

Hence proved. $\blacksquare$

Given: In $\Delta PQR$, S is on PR and T is on QR such that $\angle P = \angle RTS$.

To prove: $\Delta RPQ \sim \Delta RTS$

Proof:

In $\Delta RPQ$ and $\Delta RTS$:

1. $\angle R = \angle R$ (common angle)

2. $\angle RPQ = \angle RTS$ (given, i.e., $\angle P = \angle RTS$)

By AA similarity criterion:
$$\Delta RPQ \sim \Delta RTS$$

Hence proved. $\blacksquare$

Given: $\Delta ABE \cong \Delta ACD$.

To prove: $\Delta ADE \sim \Delta ABC$

Proof:

Since $\Delta ABE \cong \Delta ACD$ (given), their corresponding parts are equal:
$$AB = AC \quad \cdots (1)$$
$$AE = AD \quad \cdots (2)$$

Step 2: From (1) and (2):
$$\frac{AD}{AB} = \frac{AE}{AC} \quad \cdots (3)$$

(Since $AD = AE$ and $AB = AC$ means $\dfrac{AD}{AB} = \dfrac{AE}{AC}$.)

Step 3: In $\Delta ADE$ and $\Delta ABC$:

1. $\angle A = \angle A$ (common angle)

2. $\dfrac{AD}{AB} = \dfrac{AE}{AC}$ [from (3)]

By SAS similarity criterion:
$$\Delta ADE \sim \Delta ABC$$

Hence proved. $\blacksquare$

Given: AD ⊥ BC and CE ⊥ AB are altitudes of $\Delta ABC$ intersecting at P.

So $\angle ADB = \angle AEP = \angle CDB = \angle CEA = 90°$.

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(i) Prove ΔAEP ∼ ΔCDP:

In $\Delta AEP$ and $\Delta CDP$:
- $\angle AEP = \angle CDP = 90°$ (CE ⊥ AB and AD ⊥ BC)
- $\angle APE = \angle CPD$ (vertically opposite angles)

By AA similarity:
$$\Delta AEP \sim \Delta CDP \quad \blacksquare$$

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(ii) Prove ΔABD ∼ ΔCBE:

In $\Delta ABD$ and $\Delta CBE$:
- $\angle ADB = \angle CEB = 90°$ (AD ⊥ BC and CE ⊥ AB)
- $\angle ABD = \angle CBE$ (common angle B)

By AA similarity:
$$\Delta ABD \sim \Delta CBE \quad \blacksquare$$

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(iii) Prove ΔAEP ∼ ΔADB:

In $\Delta AEP$ and $\Delta ADB$:
- $\angle AEP = \angle ADB = 90°$
- $\angle A = \angle A$ (common angle at A)

By AA similarity:
$$\Delta AEP \sim \Delta ADB \quad \blacksquare$$

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(iv) Prove ΔPDC ∼ ΔBEC:

In $\Delta PDC$ and $\Delta BEC$:
- $\angle PDC = \angle BEC = 90°$ (AD ⊥ BC and CE ⊥ AB)
- $\angle PCD = \angle BCE$ (common angle C)

By AA similarity:
$$\Delta PDC \sim \Delta BEC \quad \blacksquare$$

Given: ABCD is a parallelogram. E is on AD produced. BE intersects CD at F.

To prove: $\Delta ABE \sim \Delta CFB$

Proof:

In $\Delta ABE$ and $\Delta CFB$:

1. $\angle BAE = \angle BCF$

Since ABCD is a parallelogram, AB ∥ CD. AE is a transversal (AD produced).
$\angle DAB + \angle ADC = 180°$ (co-interior angles).
Also, $\angle BAE = \angle DAB$ (same angle, since E is on AD produced, $\angle BAE = 180° - \angle BAD$... )

Actually: In parallelogram ABCD, AD ∥ BC. So $\angle AEB$ and $\angle CBF$ are alternate interior angles (since AE ∥ BC and BE is transversal).
$$\angle AEB = \angle CBF \quad \cdots (1)$$

2. $\angle ABE = \angle CFB$? Let us use:

In parallelogram ABCD: AB ∥ DC.
$\angle ABF = \angle CFB$? No.

$\angle BAE = \angle BCF$: Since AB ∥ DC (opposite sides of parallelogram), with transversal BC:
$\angle ABC + \angle BCD = 180°$.

Better approach: $\angle A = \angle C$ (opposite angles of parallelogram are equal):
$$\angle DAB = \angle BCD$$
$$\Rightarrow \angle BAE = \angle BCF \quad \cdots (1)$$ (since $\angle BAE$ is the same as $\angle DAB$ extended — actually $\angle BAE = 180° - \angle BAD$ only if E is beyond D. Let's use: since AD ∥ BC, $\angle AEB = \angle FBC$ as alternate angles.)

Cleaner proof:

In $\Delta ABE$ and $\Delta CFB$:

1. $\angle BAE = \angle BCF$
(In parallelogram, $\angle A = \angle C$, i.e., $\angle DAB = \angle DCB$; since E is on AD produced, $\angle BAE = 180° - \angle BAD$... )

Since AD ∥ BC (parallelogram), and EB is a transversal:
$$\angle AEB = \angle CBF \quad (\text{alternate interior angles}) \cdots (1)$$

2. $\angle ABE = \angle CFB$

In parallelogram AB ∥ FC (since AB ∥ DC and F is on DC):
$$\angle ABF = \angle BFC \quad (\text{alternate interior angles, AB} \parallel \text{DC}) \cdots (2)$$

From (1) and (2), by AA similarity:
$$\Delta ABE \sim \Delta CFB$$

Hence proved. $\blacksquare$

Given: $\Delta ABC$ is right-angled at B and $\Delta AMP$ is right-angled at M.

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(i) Prove ΔABC ∼ ΔAMP:

In $\Delta ABC$ and $\Delta AMP$:

1. $\angle ABC = \angle AMP = 90°$ (right angles given)

2. $\angle BAC = \angle MAP$ (common angle at A)

By AA similarity criterion:
$$\Delta ABC \sim \Delta AMP$$

Hence proved. $\blacksquare$

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(ii) Prove $\dfrac{CA}{PA} = \dfrac{BC}{MP}$:

Since $\Delta ABC \sim \Delta AMP$ (proved in part (i)), their corresponding sides are proportional:

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

In particular:
$$\frac{CA}{PA} = \frac{BC}{MP}$$

Hence proved. $\blacksquare$

Given: $\Delta ABC \sim \Delta FEG$. CD bisects $\angle ACB$ (D on AB) and GH bisects $\angle EGF$ (H on FE).

Since $\Delta ABC \sim \Delta FEG$:
$$\angle A = \angle F, \quad \angle B = \angle E, \quad \angle C = \angle G$$
$$\frac{AB}{FE} = \frac{BC}{EG} = \frac{CA}{GF} \quad \cdots (*)$$

Also, since $\angle ACB = \angle FGE$ and CD, GH are their bisectors:
$$\angle ACD = \angle DCB = \frac{\angle ACB}{2}, \quad \angle FGH = \angle HGE = \frac{\angle FGE}{2}$$
$$\Rightarrow \angle ACD = \angle FGH \quad \text{and} \quad \angle DCB = \angle HGE \quad \cdots (1)$$

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(i) Prove $\dfrac{CD}{GH} = \dfrac{AC}{FG}$:

In $\Delta ACD$ and $\Delta FGH$:
- $\angle A = \angle F$ (from similarity of $\Delta ABC$ and $\Delta FEG$)
- $\angle ACD = \angle FGH$ [from (1)]

By AA similarity: $\Delta ACD \sim \Delta FGH$

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

Hence proved. $\blacksquare$

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(ii) Prove ΔDCB ∼ ΔHGE:

In $\Delta DCB$ and $\Delta HGE$:
- $\angle B = \angle E$ (from similarity of $\Delta ABC$ and $\Delta FEG$)
- $\angle DCB = \angle HGE$ [from (1)]

By AA similarity:
$$\Delta DCB \sim \Delta HGE$$

Hence proved. $\blacksquare$

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(iii) Prove ΔDCA ∼ ΔHGF:

In $\Delta DCA$ and $\Delta HGF$:
- $\angle A = \angle F$ (from similarity of $\Delta ABC$ and $\Delta FEG$)
- $\angle DCA = \angle HGF$ [from (1), since $\angle ACD = \angle FGH$]

By AA similarity:
$$\Delta DCA \sim \Delta HGF$$

Hence proved. $\blacksquare$

Given: $\Delta ABC$ is isosceles with AB = AC. AD ⊥ BC and EF ⊥ AC. E is on CB produced.

To prove: $\Delta ABD \sim \Delta ECF$

Proof:

In $\Delta ABD$ and $\Delta ECF$:

1. $\angle ADB = \angle EFC = 90°$ (AD ⊥ BC and EF ⊥ AC)

2. Since AB = AC (isosceles triangle), the base angles are equal:
$$\angle ABC = \angle ACB$$
$$\Rightarrow \angle ABD = \angle ECF$$
(Note: $\angle ABD = \angle ABC$ and $\angle ECF = \angle ACB = \angle ECF$ since E is on BC produced, $\angle ECF = \angle ACB$)

By AA similarity criterion:
$$\Delta ABD \sim \Delta ECF$$

Hence proved. $\blacksquare$

Given: $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$, where AD and PM are medians of $\Delta ABC$ and $\Delta PQR$ respectively.

To prove: $\Delta ABC \sim \Delta PQR$

Proof:

Since AD is a median of $\Delta ABC$: $BD = \dfrac{BC}{2}$

Since PM is a median of $\Delta PQR$: $QM = \dfrac{QR}{2}$

Given: $\dfrac{AB}{PQ} = \dfrac{BC}{QR} = \dfrac{AD}{PM}$

$$\Rightarrow \frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{AD}{PM}$$

$$\Rightarrow \frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$$

In $\Delta ABD$ and $\Delta PQM$, all three sides are proportional:
$$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$$

By SSS similarity criterion:
$$\Delta ABD \sim \Delta PQM$$

Therefore:
$$\angle ABD = \angle PQM$$
$$\Rightarrow \angle ABC = \angle PQR \quad \cdots (1)$$

Now in $\Delta ABC$ and $\Delta PQR$:
- $\dfrac{AB}{PQ} = \dfrac{BC}{QR}$ (given)
- $\angle ABC = \angle PQR$ [from (1)]

By SAS similarity criterion:
$$\Delta ABC \sim \Delta PQR$$

Hence proved. $\blacksquare$

Given: In $\Delta ABC$, D is on BC such that $\angle ADC = \angle BAC$.

To prove: $CA^2 = CB \cdot CD$

Proof:

In $\Delta BAC$ and $\Delta ACD$:

1. $\angle BAC = \angle ADC$ (given)

2. $\angle ACB = \angle ACD$ (common angle C, since D is on BC)

By AA similarity criterion:
$$\Delta BAC \sim \Delta ACD$$

Therefore, corresponding sides are proportional:
$$\frac{CA}{CD} = \frac{CB}{CA}$$

$$\Rightarrow CA \times CA = CB \times CD$$

$$\Rightarrow CA^2 = CB \cdot CD$$

Hence proved. $\blacksquare$

Given: $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$, where AD and PM are medians of $\Delta ABC$ and $\Delta PQR$ respectively.

To prove: $\Delta ABC \sim \Delta PQR$

Construction: Produce AD to E such that AD = DE. Join BE and CE. Similarly, produce PM to N such that PM = MN. Join QN and RN.

Proof:

In $\Delta ABD$ and $\Delta ECD$:
- $BD = CD$ (D is mid-point of BC, as AD is median)
- $AD = ED$ (by construction)
- $\angle ADB = \angle EDC$ (vertically opposite angles)

By SAS congruence: $\Delta ABD \cong \Delta ECD$

Therefore: $AB = CE$ and $\angle ABD = \angle ECD$ (i.e., $\angle ABD = \angle ECB$, so AB ∥ CE).

Similarly, in $\Delta PQM$ and $\Delta NRM$:
$\Delta PQM \cong \Delta NRM$, so $PQ = NR$ and $PQ \parallel NR$.

Now ABEC is a parallelogram (AB ∥ CE and AB = CE), so $BE = AC$.
Similarly, PQNR is a parallelogram, so $QN = PR$.

In $\Delta ABE$: AE = 2AD (by construction), BE = AC.
In $\Delta PQN$: PN = 2PM, QN = PR.

Given: $\dfrac{AB}{PQ} = \dfrac{AC}{PR} = \dfrac{AD}{PM}$

$$\Rightarrow \frac{AB}{PQ} = \frac{BE}{QN} = \frac{AE}{PN}$$

(since $BE = AC$, $QN = PR$, $AE = 2AD$, $PN = 2PM$)

By SSS similarity: $\Delta ABE \sim \Delta PQN$

Therefore: $\angle BAE = \angle QPN$, i.e., $\angle BAC = \angle QPR$.

Now in $\Delta ABC$ and $\Delta PQR$:
- $\dfrac{AB}{PQ} = \dfrac{AC}{PR}$ (given)
- $\angle BAC = \angle QPR$ (proved above)

By SAS similarity criterion:
$$\Delta ABC \sim \Delta PQR$$

Hence proved. $\blacksquare$

Given:
- Height of pole = 6 m
- Shadow of pole = 4 m
- Shadow of tower = 28 m
- Let height of tower = $h$ m

Concept: At the same time of day, the sun's rays are parallel, so the angle of elevation of the sun is the same. Therefore, the pole and its shadow, and the tower and its shadow form similar triangles.

Setting up proportion:

Since the triangles formed are similar:
$$\frac{\text{Height of pole}}{\text{Shadow of pole}} = \frac{\text{Height of tower}}{\text{Shadow of tower}}$$

$$\frac{6}{4} = \frac{h}{28}$$

$$h = \frac{6 \times 28}{4} = \frac{168}{4} = 42 \text{ m}$$

$$\boxed{\text{Height of the tower} = 42 \text{ m}}$$

Given: $\Delta ABC \sim \Delta PQR$. AD is a median of $\Delta ABC$ (D is mid-point of BC) and PM is a median of $\Delta PQR$ (M is mid-point of QR).

To prove: $\dfrac{AB}{PQ} = \dfrac{AD}{PM}$

Proof:

Since $\Delta ABC \sim \Delta PQR$:
$$\frac{AB}{PQ} = \frac{BC}{QR} \quad \cdots (1)$$
$$\angle B = \angle Q \quad \cdots (2)$$

Since D is mid-point of BC and M is mid-point of QR:
$$BD = \frac{BC}{2} \quad \text{and} \quad QM = \frac{QR}{2}$$

From (1):
$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{\frac{BC}{2}}{\frac{QR}{2}} = \frac{BD}{QM} \quad \cdots (3)$$

In $\Delta ABD$ and $\Delta PQM$:
1. $\dfrac{AB}{PQ} = \dfrac{BD}{QM}$ [from (3)]
2. $\angle ABD = \angle PQM$ [from (2), i.e., $\angle B = \angle Q$]

By SAS similarity criterion:
$$\Delta ABD \sim \Delta PQM$$

Therefore, corresponding sides are proportional:
$$\frac{AB}{PQ} = \frac{AD}{PM}$$

Hence proved. $\blacksquare$