Exercise 6.1
Q1. Fill in the blanks using the correct word given in brackets : (i) All circles are __________. (congruent, similar) (ii) All squares are __________. (similar, congruent) (iii) All __________ triangles are similar. (isosceles, equilateral) (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
Answer:
(i) similar (All circles have same shape, not necessarily same size; thus they are similar, not necessarily congruent).
(ii) similar (All squares have same shape; corresponding angles are always 90°, and sides are proportional, so they are similar).
(iii) equilateral (All equilateral triangles have equal angles of 60° and sides in proportion, making them always similar).
(iv) (a) equal; (b) proportional (For two polygons to be similar, both conditions must hold: corresponding angles equal and corresponding sides in same ratio).
Q2. Give two different examples of pair of (i) similar figures. (ii) non-similar figures.
Answer:
(i) Similar figures:
Example 1: Two circles of different radii.
Example 2: Two squares of different side lengths.
(ii) Non-similar figures:
Example 1: A circle and a square.
Example 2: A triangle and a rectangle.
Q3. State whether the following quadrilaterals are similar or not: (Fig. 6.8 shows two quadrilaterals; one is a square and the other is a rectangle.)
Answer:
No, they are not similar. In Fig. 6.8, one quadrilateral is a square (all sides equal, all angles 90°) and the other is a rectangle (opposite sides equal, all angles 90°). Although corresponding angles are equal (90° each), corresponding sides are not in the same ratio because the ratio of sides of the square is 1:1, while the rectangle has different side lengths. Therefore, both conditions for similarity fail.
Exercise 6.2
Q1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii). [Given: Fig. 6.17 (i) shows triangle ABC with DE || BC. Side lengths: AD = 1.5 cm, DB = 3 cm, AE = 1 cm. Find EC. (ii) Triangle ABC with DE || BC. Side lengths: AD = ? (to find), DB = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm.]
Answer:
(i) In ΔABC, DE || BC. By Basic Proportionality Theorem (Theorem 6.1):
AD/DB = AE/EC
=> 1.5/3 = 1/EC
=> EC = (3 × 1)/1.5 = 2 cm
(ii) In ΔABC, DE || BC. By Theorem 6.1:
AD/DB = AE/EC
=> AD/7.2 = 1.8/5.4
=> AD = (7.2 × 1.8)/5.4 = 2.4 cm
Q2. 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
Answer:
For EF || QR, we check if PE/EQ = PF/FR (Converse of Theorem 6.2).
(i) PE/EQ = 3.9/3 = 1.3
PF/FR = 3.6/2.4 = 1.5
Since 1.3 ≠ 1.5, EF is not parallel to QR.
(ii) PE/EQ = 4/4.5 = 8/9
PF/FR = 8/9
Since 8/9 = 8/9, EF || QR.
(iii) PE = 0.18 cm, EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm
PE/EQ = 0.18/1.10 = 18/110 = 9/55
PF = 0.36 cm, FR = PR - PF = 2.56 - 0.36 = 2.20 cm
PF/FR = 0.36/2.20 = 36/220 = 9/55
Since 9/55 = 9/55, EF || QR.
Q3. In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD. [Diagram description: Fig. 6.18 shows triangle ACD with point B on AC, point M on AB extended, N on AD. Lines: LM || CB and LN || CD.]
Answer:
Proof:
In ΔABC, LM || CB (given).
By Theorem 6.1 (BPT): AM/AB = AL/AC ... (1)
In ΔACD, LN || CD (given).
By Theorem 6.1: AN/AD = AL/AC ... (2)
From (1) and (2): AM/AB = AN/AD.
Hence proved.
Q4. In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC. [Diagram description: Fig. 6.19 shows triangle ABC with points D on AB, E on BC, and F on AC. Lines: DE || AC and DF || AE.]
Answer:
Proof:
In ΔBAC, DE || AC (given).
By Theorem 6.1: BD/DA = BE/EC ... (1)
In ΔBAE, DF || AE (given).
By Theorem 6.1: BD/DA = BF/FE ... (2)
From (1) and (2): BF/FE = BE/EC.
Hence proved.
Q5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR. [Diagram description: Fig. 6.20 shows triangle OQR with point D on OQ, E on PQ extended? Actually diagram shows triangle with point P on OQ, Q on OR? The text: Points D, E, F on sides such that DE || OQ and DF || OR.]
Answer:
Proof:
In ΔPOQ, DE || OQ (given).
By Theorem 6.1: PE/EQ = PD/DO ... (1)
In ΔPOR, DF || OR (given).
By Theorem 6.1: PF/FR = PD/DO ... (2)
From (1) and (2): PE/EQ = PF/FR
Now, in ΔPQR, PE/EQ = PF/FR.
By converse of Theorem 6.2, EF || QR.
Hence proved.
Q6. 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. [Diagram description: Fig. 6.21 shows triangle OQR with point P on OQ? Actually points: O, P, Q, R with A on OP, B on OQ, C on OR. Lines: AB || PQ and AC || PR.]
Answer:
Proof:
In ΔOPQ, AB || PQ (given).
By Theorem 6.1: OA/AP = OB/BQ ... (1)
In ΔOPR, AC || PR (given).
By Theorem 6.1: OA/AP = OC/CR ... (2)
From (1) and (2): OB/BQ = OC/CR
Now, in ΔOQR, OB/BQ = OC/CR.
By converse of Theorem 6.2, BC || QR.
Hence proved.
Q7. 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. (Recall that you have proved it in Class IX).
Answer:
Given: ΔABC with D as mid-point of AB, i.e., AD = DB. Line through D parallel to BC meets AC at E.
To prove: E is mid-point of AC, i.e., AE = EC.
Proof:
In ΔABC, DE || BC (given).
By Theorem 6.1 (BPT): AD/DB = AE/EC
Since AD = DB, we have AD/DB = 1.
Therefore, AE/EC = 1, so AE = EC.
Thus, E bisects AC.
Q8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Answer:
Given: ΔABC with D as mid-point of AB and E as mid-point of AC. So AD = DB and AE = EC.
To prove: DE || BC.
Proof:
Since AD = DB, we have AD/DB = 1.
Since AE = EC, we have AE/EC = 1.
Thus, AD/DB = AE/EC.
By converse of Theorem 6.2 (Theorem 6.2), if AD/DB = AE/EC, then DE || BC.
Hence proved.
Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/CO = BO/DO.
Answer:
Given: Trapezium ABCD with AB || DC. Diagonals AC and BD intersect at O.
To prove: AO/CO = BO/DO.
Proof:
Since AB || DC, draw EF through O such that EF || AB || DC meeting AD at E and BC at F.
In ΔADC, EO || DC (by construction).
By Theorem 6.1: AE/ED = AO/OC ... (1)
In ΔABD, EO || AB.
By Theorem 6.1: AE/ED = BO/OD ... (2)
From (1) and (2): AO/OC = BO/OD, i.e., AO/CO = BO/DO.
Hence proved.
Q10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/CO = BO/DO. Show that ABCD is a trapezium.
Answer:
Given: Quadrilateral ABCD with diagonals intersecting at O, and AO/CO = BO/DO.
To prove: ABCD is a trapezium, i.e., AB || DC.
Proof:
Consider triangles AOB and COD. We have AO/CO = BO/DO (given).
Also, ∠AOB = ∠COD (vertically opposite angles).
But this is not sufficient for similarity as the ratio involves sides not including the angle.
We proceed differently:
Draw a line through O parallel to DC meeting AD at E and BC at F.
In ΔADC, EO || DC.
By Theorem 6.1: AE/ED = AO/OC ... (1)
In ΔABD, if EO || AB, then by Theorem 6.1: AE/ED = BO/OD ... (2)
Given AO/OC = BO/OD, so from (1) and (2): AE/ED = AE/ED.
Since AE/ED = AO/OC = BO/OD, and by converse of Theorem 6.2, we must have EO || AB.
Thus, EO || AB and EO || DC (by construction).
Therefore, AB || DC (since lines parallel to the same line are parallel to each other).
Hence, ABCD is a trapezium.
Activities
QActivity 1. Place a lighted bulb at a point O on the ceiling and directly below it a table in your classroom. Cut a polygon, say a quadrilateral ABCD, from a plane cardboard and place this cardboard parallel to the ground between the lighted bulb and the table. Then a shadow of ABCD is cast on the table. Mark the outline of this shadow as A'B'C'D' (see Fig.6.4). Observe that A' lies on ray OA, B' on ray OB, C' on OC and D' on OD. Verify that ∠A = ∠A', ∠B = ∠B', ∠C = ∠C', ∠D = ∠D' and AB/A'B' = BC/B'C' = CD/C'D' = DA/D'A'. This shows quadrilaterals A'B'C'D' and ABCD are similar.
Answer:
This is an activity to be performed by the student. Instructions:
1. Set up a bulb at point O on the ceiling with a table directly below.
2. Cut a quadrilateral ABCD from cardboard.
3. Hold the cardboard parallel to the ground between the bulb and table to cast a shadow.
4. Mark the shadow outline as A'B'C'D' on the table.
5. Note that A' lies on ray OA, B' on ray OB, C' on OC, D' on OD (due to light traveling in straight lines).
6. Measure the angles: ∠A = ∠A', ∠B = ∠B', ∠C = ∠C', ∠D = ∠D'.
7. Measure the sides and verify that AB/A'B' = BC/B'C' = CD/C'D' = DA/D'A'. This confirms the quadrilaterals are similar (same shape, different sizes).
QActivity 2. Draw any angle XAY and on its one arm AX, mark points (say five points) P, Q, D, R and B such that AP = PQ = QD = DR = RB. Now, through B, draw any line intersecting arm AY at C (see Fig. 6.9). Also, through the point D, draw a line parallel to BC to intersect AC at E. Observe that AD/DB = 3/2. Measure AE and EC. What about AE/EC? Observe that AE/EC is also equal to 3/2. Thus, in ΔABC, DE || BC and AD/DB = AE/EC.
Answer:
This is an activity to be performed by the student to verify the Basic Proportionality Theorem.
Steps:
1. Draw angle XAY.
2. On AX, mark points P, Q, D, R, B such that AP = PQ = QD = DR = RB. For example, mark equal segments.
3. With 5 equal segments, AD = AP + PQ + QD = 3 units, DB = DR + RB = 2 units, so AD/DB = 3/2.
4. Draw line through B meeting AY at C.
5. Through D, draw DE parallel to BC meeting AC at E.
6. Measure AE and EC using a ruler. You will find AE/EC = 3/2.
7. This verifies that when DE || BC, AD/DB = AE/EC.
QActivity 3. Draw an angle XAY on your notebook and on ray AX, mark points B1, B2, B3, B4 and B such that AB1 = B1B2 = B2B3 = B3B4 = B4B. Similarly, on ray AY, mark points C1, C2, C3, C4 and C such that AC1 = C1C2 = C2C3 = C3C4 = C4C. Then join B1C1 and BC (see Fig. 6.11). Note that AB1/B1B = AC1/C1C (Each equal to 1/4). You can see that lines B1C1 and BC are parallel to each other. Similarly, by joining B2C2, B3C3 and B4C4, observe that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side.
Answer:
This is an activity to verify the converse of the Basic Proportionality Theorem.
Steps:
1. Draw angle XAY.
2. On AX, mark 5 equal segments: AB1 = B1B2 = B2B3 = B3B4 = B4B. So AB1/AB = 1/5, AB1/B1B = 1/4.
3. On AY, mark 5 equal segments: AC1 = C1C2 = C2C3 = C3C4 = C4C. So AC1/AC = 1/5, AC1/C1C = 1/4.
4. Join B1C1. Observe that B1C1 || BC (since corresponding angles are equal due to proportion).
5. Similarly, join B2C2 (ratio 2/3), B3C3 (ratio 3/2), B4C4 (ratio 4/1) and observe each line is parallel to BC.
6. This shows: if a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
Exercises
Q1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
Answer:
In (i): Since DE || BC, by Basic Proportionality Theorem (Theorem 6.1), AD/DB = AE/EC. Given AD=1.5 cm, DB=3 cm, AE=1 cm. So 1.5/3 = 1/EC => 0.5 = 1/EC => EC = 2 cm. In (ii): Since DE || BC, AD/DB = AE/EC. Given DB=7.2 cm, AE=1.8 cm, EC=5.4 cm. So AD/7.2 = 1.8/5.4 => AD/7.2 = 1/3 => AD = 2.4 cm.
Q2. E and F are points on the sides PQ and PR respectively of a triangle 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 (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
Answer:
By converse of Basic Proportionality Theorem, EF || QR if PE/EQ = PF/FR. (i) PE/EQ = 3.9/3 = 1.3, PF/FR = 3.6/2.4 = 1.5. Since 1.3 ≠ 1.5, EF is not parallel to QR. (ii) PE/EQ = 4/4.5 = 8/9, PF/FR = 8/9. Since 8/9 = 8/9, EF || QR. (iii) Here PE = 0.18, EQ = PQ - PE = 1.28 - 0.18 = 1.1; PF = 0.36, FR = PR - PF = 2.56 - 0.36 = 2.2. PE/EQ = 0.18/1.1 = 18/110 = 9/55; PF/FR = 0.36/2.2 = 36/220 = 9/55. Since equal, EF || QR.
Q3. In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.
Answer:
In triangle ABC, LM || CB, so by Basic Proportionality Theorem, AM/AB = AL/AC. In triangle ADC, LN || CD, so AN/AD = AL/AC. Hence AM/AB = AN/AD.
Q4. In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
Answer:
In triangle ABC, DE || AC, so by Basic Proportionality Theorem, BD/DA = BE/EC. In triangle ABE, DF || AE, so BD/DA = BF/FE. Hence BF/FE = BE/EC.
Q5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
Answer:
In triangle POQ, DE || OQ, so by Basic Proportionality Theorem, PD/DO = PE/EQ. In triangle POR, DF || OR, so PD/DO = PF/FR. Thus PE/EQ = PF/FR. By converse of Basic Proportionality Theorem in triangle PQR, EF || QR.
Q6. 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.
Answer:
In triangle OPQ, AB || PQ, so OA/AP = OB/BQ. In triangle OPR, AC || PR, so OA/AP = OC/CR. Thus OB/BQ = OC/CR. In triangle OQR, by converse of Basic Proportionality Theorem, BC || QR.
Q7. 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.
Answer:
Let triangle ABC have D as midpoint of AB. Draw DE || BC meeting AC at E. By Theorem 6.1 (Basic Proportionality Theorem), AD/DB = AE/EC. Since AD = DB (D is midpoint), AD/DB = 1, so AE/EC = 1, hence AE = EC. Thus E is midpoint of AC, so the line bisects the third side.
Q8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Answer:
Let triangle ABC have D and E as midpoints of AB and AC respectively. Then AD = DB and AE = EC, so AD/DB = 1 and AE/EC = 1, hence AD/DB = AE/EC. By Theorem 6.2 (converse of Basic Proportionality Theorem), DE || BC.
Q9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/CO = BO/DO.
Answer:
In trapezium ABCD, AB || DC. In triangles AOB and COD, ∠OAB = ∠OCD (alternate interior angles) and ∠OBA = ∠ODC (alternate interior angles). Also ∠AOB = ∠COD (vertically opposite angles). Thus by AAA similarity, ∆AOB ~ ∆COD. Therefore AO/CO = BO/DO.
Q10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/CO = BO/DO. Show that ABCD is a trapezium.
Answer:
Given AO/CO = BO/DO. In triangles AOB and COD, ∠AOB = ∠COD (vertically opposite angles). The ratio of sides including these angles is equal: AO/CO = BO/DO. Thus by SAS similarity criterion, ∆AOB ~ ∆COD. Therefore ∠OAB = ∠OCD and ∠OBA = ∠ODC. Since ∠OAB = ∠OCD are alternate interior angles, AB || DC. Hence ABCD is a trapezium.
Q11. 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.
Answer:
This question refers to Fig. 6.34 which is not fully described in the text. Based on typical NCERT problems, the solution would involve checking side ratios or angle equalities. Since the figure details are missing from the text, the solution cannot be provided without the figure. The student should examine each pair and apply SSS, SAS, AA, or AAA similarity criteria as appropriate.
Q12. In Fig. 6.35, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Answer:
Given ∆ODC ~ ∆OBA, corresponding angles are equal. ∠BOC = 125° forms a linear pair with ∠DOC, so ∠DOC = 180° - 125° = 55°. In triangle ODC, ∠CDO = 70° and ∠DOC = 55°, so by angle sum property, ∠DCO = 180° - 70° - 55° = 55°. Since triangles are similar, ∠OAB = ∠DCO = 55°.
Q13. 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/OB = OC/OD.
Answer:
In trapezium ABCD, AB || DC. In triangles AOB and COD, ∠OAB = ∠OCD (alternate interior angles) and ∠OBA = ∠ODC (alternate interior angles). Also ∠AOB = ∠COD (vertically opposite angles). Thus by AAA similarity, ∆AOB ~ ∆COD. Therefore OA/OC = OB/OD, which implies OA/OB = OC/OD.
Q14. In Fig. 6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.
Answer:
Given QR/QS = QT/PR and ∠1 = ∠2. ∠1 is ∠PQS and ∠2 is ∠TQR. In triangles PQS and TQR, we have PQ/TQ = QS/QR? Actually from QR/QS = QT/PR, cross multiply: QR·PR = QS·QT. Also ∠PQS = ∠TQR. Need to verify sides including these angles. From given, we can rearrange: QR/QS = QT/PR => QR·PR = QS·QT => PR/QS = QT/QR. So sides around the equal angles: In ∆PQS, sides PQ and QS include ∠PQS; in ∆TQR, sides TQ and QR include ∠TQR. But we need PR/QS = QT/QR and ∠PQS = ∠TQR implies ∆PQS ~ ∆TQR by SAS similarity.
Q15. S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.
Answer:
In triangles RPQ and RTS, we have ∠P = ∠RTS (given). Also ∠PRQ = ∠TRS (common angle). Thus by AA similarity criterion, ∆RPQ ~ ∆RTS.
Q16. In Fig. 6.37, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC.
Answer:
Since ∆ABE ≅ ∆ACD, we have AB = AC, AE = AD, and ∠BAE = ∠CAD. From AB = AC, triangle ABC is isosceles. From AE = AD, triangle ADE is isosceles. Also ∠DAE = ∠BAC because ∠DAE = ∠BAE - ∠BAD? Actually since ∠BAE = ∠CAD, adding ∠EAD to both gives ∠BAD = ∠CAE. But to show similarity of ∆ADE and ∆ABC, note that AD/AB = AE/AC (since AD = AE and AB = AC, ratio = 1). Also ∠DAE = ∠BAC (common angle). Thus by SAS similarity, ∆ADE ~ ∆ABC.
Q17. In Fig. 6.38, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that: (i) ∆AEP ~ ∆CDP (ii) ∆ABD ~ ∆CBE (iii) ∆AEP ~ ∆ADB (iv) ∆PDC ~ ∆BEC
Answer:
(i) In ∆AEP and ∆CDP, ∠AEP = ∠CDP = 90° (altitudes). ∠APE = ∠CPD (vertically opposite angles). Thus by AA similarity, ∆AEP ~ ∆CDP. (ii) In ∆ABD and ∆CBE, ∠ADB = ∠CEB = 90° (altitudes). ∠ABD = ∠CBE (common angle). Thus by AA similarity, ∆ABD ~ ∆CBE. (iii) In ∆AEP and ∆ADB, ∠AEP = ∠ADB = 90°. ∠PAE = ∠BAD (common angle). Thus by AA similarity, ∆AEP ~ ∆ADB. (iv) In ∆PDC and ∆BEC, ∠PDC = ∠BEC = 90°. ∠PCD = ∠BCE (common angle). Thus by AA similarity, ∆PDC ~ ∆BEC.
Q18. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB.
Answer:
In parallelogram ABCD, AB || CD and AD || BC. Since E is on AD produced, we have AB || CD. In triangles ABE and CFB, ∠ABE = ∠CFB (alternate interior angles as AB || CD and BE transversal). ∠AEB = ∠CBF (alternate interior angles as AE || BC and BE transversal). Thus by AA similarity, ∆ABE ~ ∆CFB.
Q19. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that: (i) ∆ABC ~ ∆AMP (ii) CA/BC = PA/MP
Answer:
(i) In ∆ABC and ∆AMP, ∠ABC = ∠AMP = 90° (right angles). ∠BAC = ∠MAP (common angle). Thus by AA similarity, ∆ABC ~ ∆AMP. (ii) Since ∆ABC ~ ∆AMP, corresponding sides are proportional: CA/PA = BC/MP => cross multiply: CA·MP = PA·BC => CA/BC = PA/MP.
Q20. 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/AC = GH/FG (ii) ∆DCB ~ ∆HGE (iii) ∆DCA ~ ∆HGF
Answer:
Given ∆ABC ~ ∆FEG, so ∠A = ∠F, ∠B = ∠E, ∠C = ∠G. CD bisects ∠ACB so ∠ACD = ∠BCD = ∠C/2. GH bisects ∠EGF so ∠EGH = ∠FGH = ∠G/2. Since ∠C = ∠G, their halves are equal: ∠ACD = ∠FGH and ∠BCD = ∠EGH. (i) In ∆ACD and ∆FGH, ∠A = ∠F, ∠ACD = ∠FGH, so by AA, ∆ACD ~ ∆FGH, hence CD/AC = GH/FG. (ii) In ∆DCB and ∆HGE, ∠B = ∠E, ∠BCD = ∠EGH, so by AA, ∆DCB ~ ∆HGE. (iii) In ∆DCA and ∆HGF, ∠A = ∠F, ∠ACD = ∠FGH, so by AA, ∆DCA ~ ∆HGF.
Q21. 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 ∆ABD ~ ∆ECF.
Answer:
Triangle ABC is isosceles with AB = AC, so ∠ABC = ∠ACB. AD ⊥ BC so ∠ADB = 90°. EF ⊥ AC so ∠EFC = 90°. In ∆ABD and ∆ECF, ∠ABD = ∠ECF (since ∠ABC = ∠ACB and ∠ECF is same as ∠ACB). ∠ADB = ∠EFC = 90°. Thus by AA similarity, ∆ABD ~ ∆ECF.
Q22. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR. Show that ∆ABC ~ ∆PQR.
Answer:
Given AB/PQ = BC/QR = AD/PM. Since AD is median, BD = DC = BC/2. Similarly, PM is median, QM = MR = QR/2. In triangles ABD and PQM, AB/PQ = AD/PM and BD/QM = (BC/2)/(QR/2) = BC/QR. Since AB/PQ = BC/QR, we have AB/PQ = AD/PM = BD/QM. Thus by SSS similarity, ∆ABD ~ ∆PQM. Therefore ∠ABD = ∠PQM, i.e., ∠ABC = ∠PQR. Similarly, we can show another angle equal. Hence by SAS or AA, ∆ABC ~ ∆PQR.
Q23. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB·CD.
Answer:
In triangles ADC and BAC, ∠ADC = ∠BAC (given). ∠ACD = ∠BCA (common angle). Thus by AA similarity, ∆ADC ~ ∆BAC. Therefore CA/CB = CD/CA, which implies CA² = CB·CD.
Q24. 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.
Answer:
Given AB/PQ = AC/PR = AD/PM. Since AD is median, BD = DC. Similarly PM is median, QM = MR. In triangles ABD and PQM, we have AB/PQ = AD/PM and need to show BD/QM = AB/PQ. Since BD = BC/2 and QM = QR/2, we need BC/QR = AB/PQ. Not directly given but can be derived using the median property and Apollonius theorem. However, the standard approach: Extend AD to E so that AD = DE and extend PM to N so that PM = MN. Then ABEC is a parallelogram and PQNR is a parallelogram. Using given ratios, we can show ∆AEC ~ ∆PRN, leading to ∆ABC ~ ∆PQR.
Q25. 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.
Answer:
At the same time, the sun's rays make the same angle with the ground. Therefore, the triangles formed by the pole and its shadow and the tower and its shadow are similar. Let height of tower be h m. Then height of pole / shadow of pole = height of tower / shadow of tower => 6/4 = h/28 => h = (6×28)/4 = 42. So the tower is 42 m tall.
Q26. If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that AB/PQ = AD/PM.
Answer:
Given ∆ABC ~ ∆PQR, so AB/PQ = BC/QR = AC/PR. Since AD is median, BD = DC = BC/2. Since PM is median, QM = MR = QR/2. In triangles ABD and PQM, AB/PQ = BD/QM (since BD/QM = (BC/2)/(QR/2) = BC/QR = AB/PQ). Also ∠ABD = ∠PQM (corresponding angles of similar triangles). Thus by SAS similarity, ∆ABD ~ ∆PQM. Hence AB/PQ = AD/PM.