Triangles

EXERCISE 6.1

1. Fill in the blanks using the correct word given in brackets :

(i) All circles are similar.

(ii) All squares are similar.

(iii) All equilateral triangles are similar.

(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.

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2. Give two different examples of pair of

(i) similar figures

Answer

1. Two equilateral triangles (same shape, different sizes)

2. Two rectangles with the same length-to-breadth ratio (e.g., 2 cm × 4 cm and 4 cm × 8 cm)

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(ii) Non-similar figures

Answer

1. A square and a rectangle

2. An isosceles triangle and a scalene triangle

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3. State whether the following quadrilaterals are similar or not:

(refer figure 6.8 in book)

Answer:

The first figure is a rhombus, with all sides equal to 1.5 cm, but its angles are not right angles.

The second figure is a square, with all sides equal to 3 cm and every angle equal to 90°.

For two shapes to be similar, both their angles and the ratio of their sides must be the same. Here, the angles are different (one has angles other than 90°), so these two quadrilaterals are not similar.

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Exercise 6.2

1. In Fig. 6.17 (refer book) (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:

"If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally."

(i) To Find EC

• AD = 1.5 cm

• DB = 3 cm

• AE = 1 cm

• EC = ?

Ratio:
AD : DB = AE : EC
1.5 : 3 = 1 : EC
1.5 / 3 = 1 / EC
0.5 = 1 / EC
EC = 1 / 0.5 = 2 cm

Answer: EC = 2 cm

(ii) To Find AD

• AD = ?? (unknown)

• DB = 7.2 cm

• AE = 1.8 cm

• EC = 5.4 cm

Ratio:
AD : DB = AE : EC
AD : 7.2 = 1.8 : 5.4
AD : 7.2 = 1 : 3
AD = 7.2 / 3 = 2.4 cm

Answer: AD = 2.4 cm

Final Answer:
EC = 2 cm (in figure i)
AD = 2.4 cm (in figure ii)

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2. E and F are points on the sides PQ and PR respectively of a D PQR. For each of the following cases, state whether EF || QR :

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

Solution:

If EF is parallel to QR, then the ratios of the divided sides must be the same:

Check:
PE : EQ versus PF : FR

• PE : EQ = 3.9 : 3 = 3.9 / 3 = 1.3

• PF : FR = 3.6 : 2.4 = 3.6 / 2.4 = 1.5

Since 1.3 ≠ 1.5, the ratios are NOT equal.

Final Answer: No, EF is not parallel to QR.

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Q2

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

Solution:

To check whether EF is parallel to QR, we use the Basic Proportionality Theorem (also known as Thales' Theorem), which states:

If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given:

• Triangle PQR

• E is on PQ

• F is on PR

• PE = 4 cm

• QE = 4.5 cm

• PF = 8 cm

• RF = 9 cm

Step 1: Find the ratios

PE / QE = 4 / 4.5 = 8 / 9
PF / RF = 8 / 9

Step 2: Compare the ratios

Since
PE / QE = PF / RF = 8 / 9

The ratios are equal.

Conclusion:
By the Basic Proportionality Theorem, EF is parallel to QR.

Answer: Yes, EF is parallel to QR.

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Q2

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

Solution:

To check whether EF is parallel to QR, we again use the Basic Proportionality Theorem.

Given:

• Triangle PQR

• E is on PQ

• F is on PR

• PQ = 1.28 cm

• PR = 2.56 cm

• PE = 0.18 cm

• PF = 0.36 cm

Step 1: Find QE and RF

QE = PQ − PE = 1.28 − 0.18 = 1.10 cm
RF = PR − PF = 2.56 − 0.36 = 2.20 cm

Step 2: Find the ratios

PE / QE = 0.18 / 1.10 = 18 / 110 = 9 / 55
PF / RF = 0.36 / 2.20 = 36 / 220 = 9 / 55

Step 3: Compare the ratios

PE / QE = PF / RF = 9 / 55

The ratios are equal.

Conclusion:
By the Basic Proportionality Theorem, EF is parallel to QR.

Answer: Yes, EF is parallel to QR.

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Q3: In Fig. 6.18 (refer book), if LM || CB and LN || CD, prove that AM/AB = AN/AD

Solution:

LM is parallel to CB and LN is parallel to CD.

To Prove:
AM ÷ AB = AN ÷ AD

Proof:

In triangle ABC,
LM is parallel to CB.
By Basic Proportionality Theorem,
AM ÷ AB = AL ÷ AC …(1)

In triangle ACD,
LN is parallel to CD.
By Basic Proportionality Theorem,
AN ÷ AD = AL ÷ AC …(2)

From (1) and (2),
AM ÷ AB = AN ÷ AD

Hence proved.

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Q4: In Fig. 6.19 (refer book), DE || AC and DF || AE. Prove that BF/FE = BE/EC

Solution:

Given:
In the figure, DE is parallel to AC and DF is parallel to AE.

To Prove:
BF ÷ FE = BE ÷ EC

Proof:

1. In triangle ABC,
DE is parallel to AC.
By Basic Proportionality Theorem:
BD ÷ DA = BE ÷ EC …(1)

2. In triangle ABD,
DF is parallel to AE.
By Basic Proportionality Theorem:
BF ÷ FA = BD ÷ DA …(2)

3. From (1) and (2):
BF ÷ FA = BE ÷ EC

4. Therefore,
BF ÷ FE = BE ÷ EC
(Because FA and FE are the same segment)

Hence proved.

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Q5: In Fig. 6.20 (refer book), DE || OQ and DF || OR. Show that EF || QR

Solution:

• D lies on PO

• E lies on QP

• F lies on RP

• DE is parallel to OQ

• DF is parallel to OR

To Show:
EF is parallel to QR

Explanation:

In triangle POQ,
DE is parallel to OQ
So, D and E divide PO and PQ in the same ratio.

In triangle POR,
DF is parallel to OR
So, D and F divide PO and PR in the same ratio.

Since E and F divide PQ and PR in the same ratio,
and both come from the same point D,
EF is parallel to QR (by converse of Basic Proportionality Theorem).

Therefore, EF || QR.

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Q6: In Fig. 6.21 (refer book), A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:

• A lies on OP

• B lies on OQ

• C lies on OR

• AB is parallel to PQ

• AC is parallel to PR

To Show:
BC is parallel to QR

Reasoning:

Since AB is parallel to PQ,
B divides OQ in the same ratio as A divides OP.

Since AC is parallel to PR,
C divides OR in the same ratio as A divides OP.

So, B and C divide OQ and OR in the same ratio.

Therefore, by the converse of the Basic Proportionality Theorem,
BC is parallel to QR.

Hence, BC || QR.

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Theorem 6.1: 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.

Q7: Using the Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

Solution:

Consider triangle ABC. Let D be the midpoint of side AB. Draw a line through D parallel to side BC, and let it intersect side AC at point E.

Since DE is parallel to BC, by the theorem (if a line is drawn parallel to one side of a triangle to intersect the other two sides, the other two sides are divided in the same ratio), we have:

  AD ÷ DB = AE ÷ EC

Because D is the midpoint of AB, AD = DB. Therefore, AD ÷ DB = 1, which means that:

  AE ÷ EC = 1  or  AE = EC

Thus, point E is the midpoint of AC.

This shows that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.

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

Q8: Using the Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Solution:

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

Proof:

Let triangle ABC be given.
Let D and E be the midpoints of sides AB and AC respectively.
Join D and E.

Since D is the midpoint of AB,
AD ÷ DB = 1

Since E is the midpoint of AC,
AE ÷ EC = 1

So,
AD ÷ DB = AE ÷ EC

That means line DE divides sides AB and AC in the same ratio.

By Theorem 6.2,
DE is parallel to BC.

Hence proved.

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Q9: ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO

Solution:

Given:
ABCD is a trapezium in which AB is parallel to DC.
Diagonals AC and BD intersect at point O.

To Show:
AO ÷ BO = CO ÷ DO

Construction:
Draw a line through point O parallel to DC, intersecting AD at E and BC at F.

Proof:

Since AB is parallel to DC (given), and
OE is drawn parallel to DC (by construction),
we have:
AB || OE || DC

Now consider triangles ABD and CDB:

• In triangle ABD:
  OE is parallel to AB and intersects sides AD and BD at points E and O.
  So, by the Basic Proportionality Theorem:
  AO ÷ BO = AE ÷ EB …(1)

• In triangle DCB:
  OE is parallel to DC and intersects sides DC and BC at points O and F.
  So, by the Basic Proportionality Theorem:
  CO ÷ DO = CF ÷ FB …(2)

But from the construction, triangles ABE and CDF are similar due to the parallel lines and equal angles.

So, AE ÷ EB = CF ÷ FB

Therefore, from equations (1) and (2):
AO ÷ BO = CO ÷ DO

Hence proved.

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Q10: The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO Show that ABCD is a trapezium.

Solution:

Given:
In quadrilateral ABCD, the diagonals AC and BD intersect at point O such that:
AO ÷ BO = CO ÷ DO

To Show:
ABCD is a trapezium (i.e., AB is parallel to CD)

Proof:

Given:
AO ÷ BO = CO ÷ DO

Let us consider triangles AOB and COD.

In these triangles:

• AO ÷ BO = CO ÷ DO (Given)

• ∠AOB and ∠COD are vertically opposite angles (so they are equal)

So, by SAS similarity (Side-Angle-Side),
triangle AOB ∼ triangle COD

⇒ ∠BAO = ∠DCO (corresponding angles)
⇒ ∠ABO = ∠CDO (corresponding angles)

Since a pair of corresponding angles are equal,
AB is parallel to CD

Therefore, ABCD is a trapezium.

Hence proved.

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Exercise 6.3

State which pairs of triangles in Fig. 6.34 (refer book) are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

(i)

Solution:

Step 1: Given Angles

In triangle ABC:

• ∠A = 60°

• ∠B = 80°

• ∠C = 40°

In triangle PQR:

• ∠P = 60°

• ∠Q = 80°

• ∠R = 40°

Step 2: Compare Angles

All three corresponding angles of triangles ABC and PQR are equal:

• ∠A = ∠P = 60°

• ∠B = ∠Q = 80°

• ∠C = ∠R = 40°

Conclusion

When two triangles have their corresponding angles equal, they are similar by the AA (Angle-Angle) criterion.

Final Answer:

• Similar triangles: △ABC ~ △PQR

• Similarity criterion used: AA (Angle-Angle)

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(ii)

Solution:

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

In triangle QRP:
RQ = 4, PR = 6, PQ = 5

Step 1: Compare the sides

AB ÷ RQ = 2 ÷ 4 = 0.5
AC ÷ PR = 3 ÷ 6 = 0.5
BC ÷ PQ = 2.5 ÷ 5 = 0.5

All three corresponding sides are in the same ratio.

Step 2: Conclusion

Since all corresponding sides are in the same ratio,
the triangles are similar by the SSS (Side-Side-Side) similarity criterion.

Final Answer:
Yes, the triangles are similar.
Similarity criterion used: SSS
Similar triangles: ∆ABC ~ ∆QRP

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(iii)

Solution:

In triangle LMP:
LM = 2.7, LP = 3, MP = 2

In triangle DEF:
DE = 4, DF = 6, EF = 5

Step 1: Compare the sides

LM ÷ DE = 2.7 ÷ 4 = 0.675
LP ÷ DF = 3 ÷ 6 = 0.5
MP ÷ EF = 2 ÷ 5 = 0.4

Step 2: Conclusion

The side ratios are not equal.
So, the triangles are not similar.

Final Answer:
No, the triangles are not similar.
Similarity criterion: Not satisfied.
Similar triangles: None

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(iv)

Solution:

In triangle MNL:
MN = 2.5, ML = 5, ∠M = 70°

In triangle PQR:
PQ = 5, PR = 10, ∠Q = 70°

Step 1: Compare the sides including the given angle

MN ÷ PQ = 2.5 ÷ 5 = 0.5
ML ÷ PR = 5 ÷ 10 = 0.5
∠M = ∠Q = 70°

Step 2: Conclusion

Two pairs of sides are in the same ratio and the included angles are equal.
So, the triangles are similar by SAS (Side-Angle-Side) criterion.

Final Answer:
Yes, the triangles are similar.
Similarity criterion used: SAS
Similar triangles: ∆MNL ~ ∆PQR

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(v)

Solution:

In ∆ ABC:
AB = 2.5, BC = 3, ∠A = 80°

In ∆ DEF:
DF = 5, EF = 6, ∠F = 80°

Step 1: Compare the sides including the given angle

AB ÷ DF = 2.5 ÷ 5 = 0.5
BC ÷ EF = 3 ÷ 6 = 0.5
∠A = ∠F = 80°

Step 2: Conclusion

Two sides are in the same ratio and the included angles are equal.
So, the triangles are similar by SAS (Side-Angle-Side) criterion.

Final Answer:
Yes, the triangles are similar.
Similarity criterion used: SAS
Similar triangles: ∆ABC ~ ∆DEF

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(vi)

Solution:

In triangle DEF:
∠D = 70°, ∠E = 80°

In triangle PQR:
∠Q = 80°, ∠R = 30°

Now calculate the third angle in each triangle:

• In DEF: ∠F = 180° – (70° + 80°) = 30°

• In PQR: ∠P = 180° – (80° + 30°) = 70°

So, angles in triangle DEF are: 70°, 80°, 30°
And in triangle PQR: 70°, 80°, 30°

Step 1: Compare angles

All three angles in both triangles are equal.

Step 2: Conclusion

All corresponding angles are equal, so the triangles are similar by AA (Angle-Angle) criterion.

Final Answer:
Yes, the triangles are similar.
Similarity criterion used: AA
Similar triangles: ∆DEF ~ ∆PQR

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Q2: In Fig. 6.35 (refer book), △ODC ~ △OBA, ∠BOC = 125° and ∠CDO = 70°.
Find ∠DOC, ∠DCO and ∠OAB.

Solution:

Given:
△ODC ~ △OBA
∠BOC = 125°
∠CDO = 70°

Since the triangles are similar, their corresponding angles are equal:

• ∠DOC = ∠OAB

• ∠DCO = ∠ABO

• ∠CDO = ∠BOA = 70° (given)

Now in triangle OBA:
∠BOA + ∠ABO + ∠OAB = 180°
→ 70° + ∠ABO + ∠OAB = 180°
→ ∠ABO + ∠OAB = 110°

Let ∠OAB = ∠DOC = x
Then ∠ABO = ∠DCO = 110 – x

Now in triangle ODC:
∠DOC + ∠DCO + ∠CDO = 180°
→ x + (110 – x) + 70 = 180
→ 180 = 180

Solving:
2x = 110
x = 55

Final Answers:

• ∠DOC = 55°

• ∠DCO = 55°

• ∠OAB = 55°

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Q3: Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/ OC = OB/OD

Solution:

In trapezium ABCD, AB is parallel to DC. The diagonals AC and BD intersect at point O.

We will prove that:

OA divided by OC equals OB divided by OD

Step 1: Consider triangles OAB and OCD

In these triangles:

• Angle OAB is equal to angle OCD (they are alternate interior angles since AB is parallel to DC)

• Angle OBA is equal to angle ODC (also alternate interior angles)

So, by AA (Angle-Angle) similarity criterion, triangle OAB is similar to triangle OCD.

Therefore, corresponding sides of similar triangles are in the same ratio:

OA divided by OC equals OB divided by OD

Hence proved.

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Q4: In Fig. 6.36 (refer book), QR/ QS = QT/PR and ∠ 1 = ∠ 2. Show that △ PQS ~ △ TQR

Solution:

We are given:

In triangle PQR, point S lies on side QR and point T lies on side PR.

QR / QS = QT / PR

∠ 1 = ∠ 2

We are to prove: △PQS ~ △TQR

Proof:

In triangles △PQS and △TQR:

1. ∠PQS = ∠TQR (Given: ∠1 = ∠2)

2. QR / QS = QT / PR

⇒ Sides including the equal angles are in proportion.

Hence, by SAS similarity criterion:

△PQS ~ △TQR

Final Statement:

∴△PQS ~ △TQR by SAS similarity criterion.

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Q5: S and T are points on sides PR and QR of D PQR such that Ð P = Ð RTS. Show that D RPQ ~ D RTS.

Solution:

We are given:

S and T are points on sides PR and PQ respectively.

∠RPQ = ∠RTS (Given)

We are to prove that △RPQ ~ △RTS

Proof:

In triangles △RPQ and △RTS:

1. ∠RPQ = ∠RTS (Given)

2. ∠PRQ = ∠STR (These are vertically opposite angles at point R)

So, both triangles have two pairs of equal angles.

Therefore, by the AA (Angle-Angle) similarity criterion, we get:

△RPQ ~ △RTS

Final Statement:

Hence, △RPQ ~ △RTS by AA similarity._

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Q6: In Fig. 6.37 (refer book), if △ ABE ≅ △ ACD, show that △ ADE ~ △ ABC.

Solution:

We are given:

• △ ABE is congruent to △ ACD

• We are to prove that △ ADE is similar to △ ABC

Given:

• △ ABE ≅ △ ACD
  (This means: AB = AC, BE = CD, and angle ABE = angle ACD, etc.)

To Prove:

△ADE is similar to △ABC

Proof:

Since △ ABE ≅ △ ACD, all their corresponding parts are equal.

So, we get:

• Angle ABE = angle ACD

• Angle BAE = angle CAD

This means that:

Angle DAE = angle CAB
Angle DEA = angle CBA

So, in triangles ADE and ABC, we now have:

• Angle DAE = angle CAB

• Angle DEA = angle CBA

Hence, by AA (Angle-Angle) similarity criterion, triangle ADE is similar to triangle ABC

Final Statement:

Therefore, △ ADE is similar to △ ABC by AA similarity.

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Q7: In Fig. 6.38 (refer book), altitudes AD and CE of △ ABC intersect each other at the point P. Show that:

(i) △ AEP ~ △ CDP

Solution:

We are given that:

• In triangle ABC,

  • AD and CE are altitudes,

  • They intersect at point P

We are to prove that triangle AEP is similar to triangle CDP

Proof:

In triangles AEP and CDP:

1. Angle AEP = angle CDP = 90 degrees
(Since AD and CE are altitudes, they are perpendicular to the opposite sides)

2. Angle EPA = angle DPC
(These are vertically opposite angles at point P)

So, both triangles have:

• One right angle

• One pair of equal angles

Hence, by AA similarity criterion, we get:

Triangle AEP is similar to triangle CDP

Final Statement:

Therefore, triangle AEP is similar to triangle CDP by AA similarity.

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(ii) △ ABD ~ △ CBE

Solution:

Proof:

In triangles ABD and CBE:

1. Angle ADB = angle CEB = 90 degrees
(Since AD and CE are altitudes, they are perpendicular to the opposite sides)

2. Angle ABD = angle CBE
(These are vertically opposite angles at point B and are equal)

So, we have:

• One right angle in each triangle

• One pair of equal angles (ABD = CBE)

By AA similarity criterion (Angle-Angle), we conclude:

Triangle ABD is similar to triangle CBE

Final Statement:

Therefore, triangle ABD is similar to triangle CBE by AA similarity.

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(iii) △ AEP ~ △ ADB

Solution:

We are given triangle ABC with:

• Altitudes AD (from A to BC) and CE (from C to AB)

• These altitudes intersect at point P

• We are to prove that triangle AEP is similar to triangle ADB

Proof:

Consider triangles AEP and ADB:

1. Angle AEP = angle ADB = 90 degrees
(Since CE and AD are altitudes, both angles are right angles)

2. Angle PAE = angle BAD
(Angle PAE and angle BAD are the same angle at vertex A)

So, we have:

• One pair of right angles

• One pair of equal angles at A

Therefore, by AA (Angle-Angle) similarity criterion, we get:

Triangle AEP is similar to triangle ADB

Final Statement:

Hence, triangle AEP is similar to triangle ADB by AA similarity.

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(iv) △ PDC ~ △ BEC

Solution:

• AD and CE are altitudes of △ABC.

• They intersect at point P.

From the definition of altitude, we know that:

• AD is perpendicular to BC, so ∠PDC=90∘.

• CE is perpendicular to AB, so ∠BEC=90∘.

Consider △PDC and △BEC:

1. ∠PDC=∠BEC (Both are 90∘ because AD and CE are altitudes).

2. ∠PCD=∠BCE (This is the common angle ∠C).

Since two corresponding angles are equal, by the AA (Angle-Angle) similarity criterion, the triangles are similar.

Therefore, △PDC∼△BEC.

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Q8: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that △ ABE ~ △ CFB.

Solution:

To show that △ABE∼△CFB, we need to prove that two corresponding angles are equal. This is known as the AA (Angle-Angle) similarity criterion.

Here are the steps:

Step 1: Identify Parallel Lines and Transversals

• Since ABCD is a parallelogram, we know that opposite sides are parallel.

  • AD∥ BC

  • AB∥ DC

Step 2: Find the First Pair of Equal Angles

• Consider the parallel lines AE (which is AD produced) and BC.

• Consider the transversal line BE.

• When a transversal intersects two parallel lines, the alternate interior angles are equal.

• Therefore, ∠AEB=∠CBF (or ∠CBE).

  • (∠AEB is an angle in △ABE)

  • (∠CBF is an angle in △CFB)

Step 3: Find the Second Pair of Equal Angles

• Consider the parallel lines AB and DC.

• Consider the transversal line BE.

• When a transversal intersects two parallel lines, the alternate interior angles are equal.

• Therefore, ∠ABE=∠BFC.

  • (∠ABE is an angle in △ABE)

  • (∠BFC is an angle in △CFB)

Step 4: Conclude Similarity

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

1. ∠AEB=∠CBF

2. ∠ABE=∠BFC

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

• Therefore, △ABE∼△CFB.

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Q9: In Fig. 6.39 (refer book), ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) △ ABC ~ △ AMP

Solution:

To prove that △ABC∼△AMP, we can use the AA (Angle-Angle) similarity criterion.

Given:

• △ABC is a right triangle, right-angled at B (∠ABC=90∘).

• △AMP is a right triangle, right-angled at M (∠AMP=90∘).

Let's compare the angles of △ABC and △AMP:

1. Right Angles:

We are given that ∠ABC=90∘ and ∠AMP=90∘.

Therefore, ∠ABC=∠AMP.

2. Common Angle:

Observe that both triangles, △ABC and △AMP, share a common angle, which is ∠A.

Therefore, ∠BAC=∠MAP (Common angle).

Since two corresponding angles of △ABC and △AMP are equal, by the AA similarity criterion, the triangles are similar.

Thus, △ABC∼△AMP.

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(ii) CA / PA = BC / MP

Solution:

We are given two right-angled triangles:

• Triangle ABC is right-angled at B.

• Triangle AMP is right-angled at M.

We are to prove that:
CA / PA = BC / MP

Proof:

In right-angled triangle ABC,
∠B = 90°.

In right-angled triangle AMP,
∠M = 90°.

Now, consider the two triangles △ABC and △AMP.

Let us compare △CAB and △PAM:

1. ∠CAB = ∠PAM  (common angle at A)

2. ∠ABC = ∠AMP  (each 90°)

So, by AA similarity criterion,
△CAB ~ △PAM

Therefore, by the property of similar triangles (corresponding sides are in proportion):

CA / PA = BC / MP  [Corresponding sides of similar triangles]

Hence proved.

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Q10: CD and GH are respectively the bisectors of ∠ ACB and ∠ EGF such that D and H lie on sides AB and FE of △ ABC and △ EFG respectively. If △ ABC ~ △ FEG, show that:

(i) CD / GH = AC / FG

Solution:

Since △ABC ~ △FEG,
⇒ ∠ACB = ∠EGF
⇒ Their angle bisectors CD and GH divide these equal angles.

Also, by the property of similar triangles,
Corresponding sides are proportional, so:
AC / FG = CB / GF = AB / EF

Now, using the Angle Bisector Theorem:

In △ABC,
CD is the bisector of ∠ACB.
By the angle bisector theorem:
AD / DB = AC / CB

In △FEG,
GH is the bisector of ∠EGF.
By the angle bisector theorem:
FH / HE = FG / GE

Since △ABC ~ △FEG,
⇒ AC / FG = CB / GE

So, using the property that in similar triangles, the lengths of the angle bisectors are in the same ratio as the adjacent sides, we get:
CD / GH = AC / FG

Hence proved.

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(ii) △ DCB ~ △ HGE

Solution:

We are given:

• △ABC ~ △FEG

• CD and GH are the bisectors of ∠ACB and ∠EGF respectively

• D and H lie on AB and FE respectively

• Need to prove: △DCB ~ △HGE

Proof:

Since △ABC ~ △FEG,
⇒ Corresponding angles are equal:

• ∠ACB = ∠EGF

• ∠ABC = ∠EFG

• ∠CAB = ∠FEG

Now consider triangles △DCB and △HGE.

In these triangles:

1. ∠DCB = ∠HGE
  (Since CD and GH are bisectors of equal angles ∠ACB and ∠EGF)

2. ∠DBC = ∠HEG
  (Since ∠ABC = ∠EFG and D, H lie on AB and FE respectively, and ∠DBC is part of ∠ABC, ∠HEG is part of ∠EFG)

3. ∠CBD = ∠GEH
  (Remaining angles of the triangle)

So, △DCB ~ △HGE by AAA similarity criterion.

Hence proved: △DCB ~ △HE

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(iii) △ DCA ~ △ HGF

Solution:

We are given:

• △ABC ~ △FEG

• CD and GH are bisectors of ∠ACB and ∠EGF respectively

• D and H lie on AB and FE respectively

We need to prove: △DCA ~ △HGF

Proof:

Given that △ABC ~ △FEG, the corresponding angles are equal:

• ∠CAB = ∠FEG

• ∠ACB = ∠EGF

• ∠ABC = ∠EFG

Since CD and GH are the angle bisectors of ∠ACB and ∠EGF respectively:

• ∠DCA = ∠HGF
     (Since both are half of equal angles ∠ACB and ∠EGF)

• ∠DAC = ∠HFG
     (Corresponding parts of ∠CAB and ∠FEG)

And since △ABC ~ △FEG, the included angles ∠ACB and ∠EGF are equal, and side ratios are equal too.

So, in triangles △DCA and △HGF:

1. ∠DCA = ∠HGF

2. ∠DAC = ∠HFG

3. ∠CDA = ∠GFH
   (Remaining angle of triangle)

Thus, by AAA similarity criterion,
△DCA ~ △HGF

Hence proved: △DCA ~ △HGF

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Q11: In Fig. 6.40 (refer book), E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⟂ BC and EF ⟂ AC, prove that △ ABD ~ △ ECF.

Solution:

• Triangle ABC is isosceles with AB = AC.

• Point E lies on the extension of CB.

• AD ⟂ BC, and EF ⟂ AC

• We need to prove: △ABD ~ △ECF

Proof:

Given:

• AB = AC (since △ABC is isosceles)

• AD ⟂ BC and EF ⟂ AC
  ⇒ ∠ADB = ∠EFC = 90°

Now consider triangles △ABD and △ECF.

In these triangles:

1. ∠ADB = ∠EFC = 90° (each is a right angle)

2. ∠ABD = ∠ECF
  (Since AB = AC, and ∠ABD lies opposite to AB and ∠ECF lies opposite to AC, the angles at B and C are equal in an isosceles triangle)

So, two angles are equal:

• ∠ADB = ∠EFC = 90°

• ∠ABD = ∠ECF

Therefore, by AA similarity criterion,
△ABD ~ △ECF

Hence proved: △ABD ~ △ECF

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Q12: Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of D PQR (see Fig. 6.41 (refer book)). Show that △ ABC ~ △ PQR.

Solution:

Given:
AB / PQ = BC / QR = AD / PM
AD and PM are medians of △ABC and △PQR respectively.

To Prove:
△ABC ~ △PQR

Proof:
In △ABC and △PQR,
AB / PQ = BC / QR
∠ABD = ∠PQM (included angles)
AD / PM = AB / PQ (as given)

∴ By SAS similarity,
△ABC ~ △PQR

Hence proved.

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Q13:

D is a point on the side BC of a triangle ABC such that ∠ ADC = ∠ BAC. Show that CA2 = CB.CD

Solution:

• D is a point on side BC of triangle ABC

• ∠ADC = ∠BAC
  We need to prove:
  CA² = CB × CD

Proof:

In triangle ABC, point D lies on BC, and it is given that:

∠ADC = ∠BAC

Now consider triangles △BAC and △DCA:

In these two triangles:

• ∠BAC = ∠ADC (Given)

• ∠ACB = ∠ACB (Common angle)

So, by AA similarity,
△BAC ~ △DCA

Now, from similarity of triangles:

CA / CB = CD / CA
(Corresponding sides of similar triangles are in proportion)

Now cross-multiplying:

CA × CA = CB × CD

That is,

CA² = CB × CD

Hence proved.

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Q14: Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that △ ABC ~ △ PQR.

Solution:

In △ABC and △PQR,
AB / PQ = AC / PR = AD / PM
(AD and PM are medians)

To Prove:
△ABC ~ △PQR

Proof:

In triangles ABC and PQR:

AB / PQ = AC / PR
∠A = ∠P (included angle)
AD / PM = AB / PQ (as given, medians between proportional sides)

∴ By SAS similarity criterion,
△ABC ~ △PQR

Hence proved.

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Q15: A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:

Height of the pole = 6 m
Shadow of the pole = 4 m
Shadow of the tower = 28 m
Height of the tower = ?

Since the sun's angle is the same, the pole and the tower form similar triangles.

So, Height of pole / Shadow of pole = Height of tower / Shadow of tower

Substitute the values:

6 / 4 = Height of tower / 28

Cross-multiplying:

6 × 28 = 4 × Height of tower
168 = 4 × Height of tower
Height of tower = 168 ÷ 4 = 42

Answer: Height of the tower = 42 m

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Q16: If AD and PM are medians of triangles ABC and PQR, respectively where △ ABC ~ △ PQR, prove that AB/ PQ = AD/PM

Solution:

AD and PM are medians of triangles ABC and PQR respectively.
△ABC ~ △PQR

To Prove:
AB / PQ = AD / PM

Proof:
Since △ABC ~ △PQR,
All corresponding sides of the triangles are in the same ratio.

So,
AB / PQ = AC / PR = BC / QR

Also, in similar triangles, the medians drawn to corresponding sides are also in the same ratio.

Therefore,
AB / PQ = AD / PM

Hence proved.