Chapter 6 - Triangles

Exercise 6.1

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

  1. All circles are _______. (congruent, similar)
  2. All squares are _______. (similar, congruent)
  3. All triangles are _______ similar. (isosceles, equilateral)
  4. Two polygons of the same number of sides are similar, if
    1. their corresponding angles are _______ and
    2. their corresponding sides are _______. (equal, proportional)

  1. (i) All circles are similar.

    (ii) All squares are similar.

    (iii) All triangles are not similar (only if equilateral then yes). (Here the intended answer is “equilateral” triangles are always similar.)

    (iv) Two polygons of the same number of sides are similar if:

    1. their corresponding angles are equal, and
    2. their corresponding sides are proportional.

2. Give two different examples of pair of

  1. similar figures.
  2. non-similar figures.

  • (i) Examples of similar figures: two circles of different sizes, two squares of different sizes.

    (ii) Examples of non-similar figures: a circle and a square, a rectangle and a triangle.

    • 3. State whether the following quadrilaterals are similar or not:
    Quadrilaterals for similarity analysis
    The answer is no, because their corresponding sides are proportional (they have the same ratio), but their corresponding angles are not equal.

    Exercise 6.2

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

    Quadrilaterals for similarity analysis

    Apply the Basic Proportionality Theorem (BPT):

    If a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, then the line divides those two sides in the same ratio.

    In this case, DE || BC. Therefore, by BPT, AD/DB = AE/EC.

    Substitute the given values:

    So, 1.5 / 3 = 1 / EC.

    Solve for EC:

    Multiply both sides by EC to get: 1.5 × EC = 3.

    Divide both sides by 1.5: EC = 3 / 1.5.

    EC = 2 cm.

    Conclusion: Therefore, EC = 2 cm.

    2. E and F are points on the sides PQ and PR respectively of a ΔPQR. 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

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

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

    Solution:

    Identify the given values:

    Calculate the ratio of PE to EQ:

    PE / EQ = 4 / 4.5 = 8 / 9

    Calculate the ratio of PF to FR:

    PF / FR = 8 / 9

    Compare the ratios:

    PE / EQ = PF / FR = 8 / 9

    Apply the converse of the Basic Proportionality Theorem:

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

    3. In Fig. 6.18, if LM || CB and LN || CD, prove that
    AM/AN = AB/AD

    Solution:

    In the diagram, if we have a triangle ABC and a line DE such that D lies on AB and E lies on AC, and AD/DB = AE/EC (or AD/AB = AE/AC), then we can use the Basic Proportionality Theorem (BPT) to prove that DE is parallel to BC.

    1. Basic Proportionality Theorem: The BPT states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, it will divide those sides in the same ratio.
    2. The 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.
    3. Applying the converse of BPT: Since we are given that AD/DB = AE/EC, we can use the converse of the BPT to conclude that DE is parallel to BC.
    4. Therefore: Line DE is parallel to line BC.

    4. In Fig. 6.19, DE || AC and DF || AE. Prove that
    BF/BE = FE/EC

    Given Information: In triangle PQR, points E and F are on sides PQ and PR respectively.

    Applying the Converse of Basic Proportionality Theorem: To prove that EF is parallel to QR, we need to show that PE/EQ = PF/RF.

    Calculating Ratios:

    Comparing Ratios: Since 1.3 ≠ 1.5, the ratios are not equal.

    Conclusion: According to the Converse of Basic Proportionality Theorem, EF is not parallel to QR.

    Therefore, the answer to question 4 in Exercise 6.2 is that EF is not parallel to QR.

    5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.

    Given:

    To prove: EF || QR

    Proof:

    1. Since DE || OQ and both intersect ∆POQ, by the Basic Proportionality Theorem (BPT), we get:
      PE / EQ = PD / DO
    2. Since DF || OR and both intersect ∆POR, by BPT again:
      PF / FR = PD / DO
    3. From both equations:
      PE / EQ = PF / FR
    4. Thus, the line segment EF divides PQ and PR in the same ratio.
    5. By the **Converse of Basic Proportionality Theorem**, this implies:
      EF || QR

    Conclusion: Hence, EF is parallel to QR.

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

    Proof:

    1. AB || PQ implies: OA / AP = OB / BQ (By BPT)
    2. AC || PR implies: OA / AP = OC / CR (By BPT)
    3. From (1) and (2): OB / BQ = OC / CR
    4. So, BC divides QR in the same ratio.
    5. Therefore, by Converse of BPT: BC || QR

    Hence proved: BC || QR

    7. Using 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.

    Given: Triangle ABC with D as midpoint of AB and line DE || BC.

    To Prove: E is midpoint of AC.

    1. Since DE || BC, by Basic Proportionality Theorem:
    2. AD / DB = AE / EC
    3. AD = DB (since D is midpoint), so 1 = AE / EC
    4. Hence, AE = EC. So E is midpoint of AC.

    Therefore: A line through midpoint and parallel to another side bisects the third side.

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

    Given: Triangle ABC with D and E as midpoints of AB and AC respectively.

    To Prove: DE || BC

    1. D and E are midpoints of AB and AC.
    2. Then by Midpoint Theorem (Theorem 6.2):
    3. The line joining midpoints of two sides is parallel to the third side.
    4. Hence, DE || BC

    Therefore: DE is parallel to BC.

    9. ABCD is a trapezium where AB || DC and diagonals intersect at point O. Show that:

    AO / CO = BO / DO

    Proof:

    1. AB || DC and diagonals AC and BD intersect at O.
    2. Triangles AOB and COD are formed.
    3. By Basic Proportionality Theorem in both triangles:
    4. In ∆AOB and ∆COD, AB || DC ⇒ corresponding sides proportional
    5. AO / CO = BO / DO

    Hence proved.

    10. The diagonals of quadrilateral ABCD intersect at O such that:

    AO / CO = BO / DO

    Show that ABCD is a trapezium.

    Proof:

    1. Given: AO / CO = BO / DO
    2. In triangle AOB and COD, the intersecting diagonals divide each other in the same ratio.
    3. By converse of Basic Proportionality Theorem:
    4. Lines AB and DC must be parallel.
    5. Hence, ABCD is a trapezium (only one pair of opposite sides are parallel).

    Therefore: ABCD is a trapezium.

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

    Proof:

    1. AB || PQ implies: OA / AP = OB / BQ (By BPT)
    2. AC || PR implies: OA / AP = OC / CR (By BPT)
    3. From (1) and (2): OB / BQ = OC / CR
    4. So, BC divides QR in the same ratio.
    5. Therefore, by Converse of BPT: BC || QR

    Hence proved: BC || QR

    7. Using 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.

    Given: Triangle ABC with D as midpoint of AB and line DE || BC.

    To Prove: E is midpoint of AC.

    1. Since DE || BC, by Basic Proportionality Theorem:
    2. AD / DB = AE / EC
    3. AD = DB (since D is midpoint), so 1 = AE / EC
    4. Hence, AE = EC. So E is midpoint of AC.

    Therefore: A line through midpoint and parallel to another side bisects the third side.

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

    Given: Triangle ABC with D and E as midpoints of AB and AC respectively.

    To Prove: DE || BC

    1. D and E are midpoints of AB and AC.
    2. Then by Midpoint Theorem (Theorem 6.2):
    3. The line joining midpoints of two sides is parallel to the third side.
    4. Hence, DE || BC

    Therefore: DE is parallel to BC.

    9. ABCD is a trapezium where AB || DC and diagonals intersect at point O. Show that:

    AO / CO = BO / DO

    Proof:

    1. AB || DC and diagonals AC and BD intersect at O.
    2. Triangles AOB and COD are formed.
    3. By Basic Proportionality Theorem in both triangles:
    4. In ∆AOB and ∆COD, AB || DC ⇒ corresponding sides proportional
    5. AO / CO = BO / DO

    Hence proved.

    10. The diagonals of quadrilateral ABCD intersect at O such that:

    AO / CO = BO / DO

    Show that ABCD is a trapezium.

    Proof:

    1. Given: AO / CO = BO / DO
    2. In triangle AOB and COD, the intersecting diagonals divide each other in the same ratio.
    3. By converse of Basic Proportionality Theorem:
    4. Lines AB and DC must be parallel.
    5. Hence, ABCD is a trapezium (only one pair of opposite sides are parallel).

    Therefore: ABCD is a trapezium.

    Exercise 6.3

    1. State which pairs of triangles in Fig. 6.34 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 :

    Understanding Similarity Criteria:

    Examples from Fig. 6.34:

    Example (i):
    In ΔABC and ΔPQR:
    ∠A = ∠P = 60°, ∠B = ∠Q = 80°, and ∠C = ∠R = 40°
    All corresponding angles are equal.
    ΔABC ~ ΔPQR by AAA similarity.

    Example (ii):
    In ΔABC and ΔPQR:
    AB / QR = 2 / 4 = 1/2
    BC / RP = 2.5 / 5 = 1/2
    AC / PQ = 3 / 6 = 1/2
    All sides are in proportion.
    ΔABC ~ ΔPQR by SSS similarity.

    2. In Fig. 6.35, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OCD.

    Given: ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°

    To find: ∠DOC, ∠DCO, and ∠OCD

    Solution:

    Since ΔODC ~ ΔOBA, we know that:

    ∠DOC = ∠BOA (corresponding angles)

    ∠DCO = ∠OAB (corresponding angles)

    ∠OCD = ∠OBA (corresponding angles)

    We also know that ∠BOC = 125° and ∠CDO = 70°.

    To find ∠DOC, we can use the fact that the sum of angles in a triangle is 180°.

    ∠DOC = 180° - ∠BOC - ∠DCO

    ∠DOC = 180° - 125° - 70°

    ∠DOC = 180° - 195°

    ∠DOC = 5°

    Therefore, ∠DOC = 5°.

    Similarly, we can find ∠DCO and ∠OCD.

    ∠DCO = 180° - ∠BOC - ∠DOC

    ∠DCO = 180° - 125° - 5°

    ∠DCO = 50°

    ∠OCD = 180° - ∠BOC - ∠DCO

    ∠OCD = 180° - 125° - 50°

    ∠OCD = 5°

    Therefore, ∠DOC = 5°, ∠DCO = 50°, and ∠OCD = 5°.

    3. 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

    Given: Diagonals of trapezium ABCD intersect at O, with AB || DC.

    To prove: OA/OC = OB/OD

    Proof:

    1. In ΔAOB and ΔCOD, AB || DC ⇒ corresponding angles are equal.
    2. ∠AOB = ∠COD (corresponding angles)
    3. ∠OAB = ∠OCD (corresponding angles)
    4. ∠OBA = ∠ODC (corresponding angles)
    5. By AAA similarity criterion, ΔAOB ~ ΔCOD
    6. ⇒ OA/OC = OB/OD (corresponding sides are proportional)

    Therefore: OA/OC = OB/OD

    4. In Fig. 6.36

    QR / QT = QS / PR and ∠1 = ∠2.
    Show that ΔPQS ~ ΔTQR.

    Solution:

    We are given the following:

    Now consider triangles PQS and TQR:

    1. The sides are in the same ratio: QR/QT = QS/PR.
    2. The included angles are equal: ∠1 = ∠2.

    So, the triangles have one angle equal and the sides including those angles in the same ratio.

    Therefore, by SAS (Side-Angle-Side) similarity criterion:

    ΔPQS ~ ΔTQR

    5. S and T are points on sides PR and QR of triangle PQR

    It is given that ∠P = ∠RTS.
    Show that ΔRPQ ~ ΔRTS.

    Solution:

    Given: ∠P = ∠RTS, and S and T are points on sides PR and QR respectively of triangle PQR.

    Consider triangles RPQ and RTS.

    1. Angle ∠P in triangle RPQ is equal to angle ∠RTS in triangle RTS (given).
    2. Angle ∠RPQ and ∠TRS are the same angle (common angle).

    So, both triangles have two corresponding angles equal.

    Therefore, by AAA (Angle-Angle-Angle) similarity criterion:

    ΔRPQ ~ ΔRTS

    6. In Fig. 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

    Given: ΔABE ≅ ΔACD

    To prove: ΔADE ~ ΔABC

    Proof:

    1. Given: ΔABE ≅ ΔACD
    2. ⇒ AB = AC (corresponding sides)
    3. ∠BAE = ∠CAD (corresponding angles)
    4. ∠AEB = ∠ADC (corresponding angles)
    5. By SAS similarity criterion, ΔADE ~ ΔABC

    Therefore: ΔADE ~ ΔABC

    7. In Fig. 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

    (i) ΔAEP ~ ΔCDP

    (ii) ΔABD ~ ΔCBE

    (iii) ΔAEP ~ ΔADB

    (iv) ΔPDC ~ ΔBEC

    Explanation:

    We know that if two triangles are congruent, then their corresponding parts are equal.

    If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the triangles are similar. This is known as the SAS (Side-Angle-Side) similarity criterion.

    In ∆ABE and ∆ACD:

    Now, consider ∆ADE and ∆ABC:

    Therefore, by the SAS similarity criterion:

    ∆ADE ~ ∆ABC

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

    Solution:

    We are given that ABCD is a parallelogram and E is a point on the extension of side AD. Line BE intersects CD at F.

    To prove: ∆ABE ~ ∆CFB

    Proof:

    Now, in ∆ABE and ∆CFB:

    So, by AA (Angle-Angle) similarity criterion:

    ∆ABE ~ ∆CFB

    9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

    1. ∆ABC is similar to ∆AMP.
    2. CA/PA = BC/MP.

    (i)

    If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This is known as the AA (Angle-Angle) similarity criterion.

    In triangles ΔABC and ΔAMP:

    Therefore, by the AA criterion,
    ΔABC ∼ ΔAMP

    (ii)

    Since the triangles are similar, the ratios of their corresponding sides are equal.

    Thus, in ΔABC and ΔAMP:
    CA / PA = BC / MP

    10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF and ∠BDC = 50°. Find ∠HGE.

    (i) CD divided by AC equals GH divided by FG.

    (ii)Triangle DCB is similar to triangle HGE.

    (iii)Triangle DCA is similar to triangle HGF.

    (iv)Triangle DCA is similar to triangle HGF.

    We have angle bisectors:

    given statement

    Since CD and GH are angle bisectors, they divide their respective angles into two equal parts. When we have angle bisectors, they create similar triangles.

    Given that ∠BDC = 50°, and triangle DCB is similar to triangle HGE (from statement ii), this means that ∠HGE would also be 50°.

    Therefore, ∠HGE = 50° is the answer to this geometry question

    In Fig. 6.40, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

    Answer:

    Given:

    To Prove: ΔABD ~ ΔECF

    Proof:

    1. ∠ADB = 90° (AD ⊥ BC)
    2. ∠EFC = 90° (EF ⊥ AC)
    3. ∠BAD = ∠EFC (Common angle at vertex A)
    4. By AA similarity criterion, ΔABD ~ ΔECF

    Question 12

    In triangle ABC, sides AB, BC, and median AD are respectively proportional to sides PQ, QR, and median PM of another triangle PQR. Show that ΔABC ∼ ΔPQR.

    Solution:

    Given:

    To prove: ΔABC ∼ ΔPQR

    Proof:

    1. Given proportionality: AB/PQ = BC/QR = AD/PM
    2. Construct points E on AD such that AD = DE, and N on PM such that PM = MN.
    3. In triangles ABD and ECD:
      • BD = CD (D is midpoint of BC)
      • AD = DE (by construction)
      • ∠ADB = ∠CDE (vertically opposite angles)
      • Therefore, by SAS congruence, ΔABD ≅ ΔECD, so AB = EC.
    4. In triangles PQM and PRM:
      • QM = RM (M is midpoint of QR)
      • PM = MN (by construction)
      • ∠QPM = ∠RPM (vertically opposite angles)
      • Therefore, by SAS congruence, ΔPQM ≅ ΔPRM, so PQ = NR.
    5. From the congruences: EC = AB and NR = PQ, so EC/AB = NR/PQ = 1.
    6. Therefore, by SAS similarity criterion, ΔABC ∼ ΔPQR.
    13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB × CD.

    Solution:

    Given: D lies on BC such that ∠ADC = ∠BAC.

    To prove: CA squared equals CB multiplied by CD, i.e., CA² = CB × CD.

    Proof:
    - Consider triangles ADC and BAC.
    - Given ∠ADC = ∠BAC.
    - Also, ∠ACD = ∠ABC (because they are alternate angles when AD is a transversal).
    - Therefore, by AA similarity criterion, ΔADC ~ ΔBAC.

    Since the triangles are similar:
    CA / BA = CD / BA = AC / AB = DC / CB (corresponding sides are proportional)

    Focusing on sides:
    CA / CB = CD / CA
    Cross-multiplied:
    CA × CA = CB × CD
    Or,
    CA² = CB × CD.

    Hence proved.
    14. 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 ΔABC ~ ΔPQR.

    Solution:

    Given:
    AB / PQ = AC / PR = AD / PM.

    To prove: ΔABC ~ ΔPQR.

    Proof:
    - Since medians AD and PM correspond in the two triangles,
    - Construct points and use median properties:
    - In triangles ABD and PQM,
    - AB / PQ = AD / PM (given proportionality)
    - BD = QM (since D and M are midpoints of BC and QR)
    - ∠ADB = ∠PMQ (vertically opposite angles)
    - By SAS criterion, ΔABD ~ ΔPQM.
    - Similarly, in triangles ACD and PRM,
    - AC / PR = AD / PM (given)
    - CD = MR (midpoints property)
    - ∠ADC = ∠PMR (vertically opposite angles)
    - By SAS criterion, ΔACD ~ ΔPRM.

    - Since ABD and ACD together make up ABC and PQM and PRM make up PQR,
    - ΔABC ~ ΔPQR.

    Hence proved.
    15. 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:

    Given:
    - Height of pole = 6 m
    - Shadow of pole = 4 m
    - Shadow of tower = 28 m
    - Height of tower = ?

    Assumption:
    - The sun rays make the same angle with the ground for both the pole and the tower.
    - Hence, triangles formed by the pole and its shadow and tower and its shadow are similar.

    Applying similarity of triangles:
    Height of pole / Shadow of pole = Height of tower / Shadow of tower
    6 / 4 = Height of tower / 28

    Cross multiply:
    Height of tower = (6 × 28) / 4 = 168 / 4 = 42 m

    Therefore, the height of the tower is 42 meters.
    16. If AD and PM are medians of triangles ABC and PQR respectively where ΔABC ~ ΔPQR, prove that AB / AD = PQ / PM.

    Solution:

    Given:
    - AD and PM are medians of triangles ABC and PQR respectively.
    - ΔABC ~ ΔPQR.

    To prove:
    AB / AD = PQ / PM.

    Proof:
    - Since ΔABC ~ ΔPQR, corresponding sides are proportional:
    AB / PQ = BC / QR = AC / PR = k (some constant ratio).
    - Also, corresponding medians of similar triangles are proportional to their corresponding sides.

    - By properties of medians in similar triangles:
    AD / PM = AB / PQ = k

    - Rearranging:
    AB / AD = PQ / PM.

    Hence proved that the ratio of a side to the median drawn to the opposite side in one triangle is equal to the ratio of the corresponding side to the corresponding median in the similar triangle.