Triangles - NCERT Solutions

Exercise 6.1
Exercise 6.2
Exercise 6.3
Theorems & Examples

Exercise 6.1

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

(i) All circles are similar (congruent, similar)

(ii) All squares are similar (similar, congruent)

(iii) All equilateral triangles are similar. (isosceles, equilateral)

(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 (equal, proportional)

2. Give two different examples of pair of

(i) Similar figures:

  • Any two circles
  • Any two squares

(ii) Non-similar figures:

  • A square and a rectangle (unless the rectangle is also a square)
  • A circle and a triangle
3. State whether the following quadrilaterals are similar or not:
[Refer to Fig. 6.8 in the textbook]
Fig. 6.8

The quadrilaterals are not similar because:

  • Corresponding angles are equal (all are 90°)
  • But corresponding sides are not proportional (3/1.5 = 2, but 3/3 = 1)

Exercise 6.2

1. In Fig. 6.17, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
[Refer to Fig. 6.17 in the textbook]
Fig. 6.17

(i) Finding EC:

Using Basic Proportionality Theorem (Thales Theorem):

AD/DB = AE/EC

1.5/3 = 1/EC

EC = (3 × 1)/1.5 = 2 cm

(ii) Finding AD:

Using Basic Proportionality Theorem:

AD/DB = AE/EC

AD/7.2 = 1.8/5.4

AD = (1.8 × 7.2)/5.4 = 2.4 cm

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

PE/EQ = 3.9/3 = 1.3

PF/FR = 3.6/2.4 = 1.5

Since PE/EQ ≠ PF/FR, EF is not parallel to QR.

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

PE/EQ = 4/4.5 = 8/9

PF/FR = 8/9

Since PE/EQ = PF/FR, EF is parallel to QR.

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

PE/EQ = 0.18/(1.28-0.18) = 0.18/1.10 = 9/55

PF/FR = 0.36/(2.56-0.36) = 0.36/2.20 = 9/55

Since PE/EQ = PF/FR, EF is parallel to QR.

3. In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD
[Refer to Fig. 6.18 in the textbook]
Fig. 6.18

Proof:

In ΔABC, LM || CB (Given)

By Basic Proportionality Theorem: AM/AB = AL/AC ...(1)

In ΔADC, LN || CD (Given)

By Basic Proportionality Theorem: AN/AD = AL/AC ...(2)

From (1) and (2): AM/AB = AN/AD

Hence proved.

4. In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC
[Refer to Fig. 6.19 in the textbook]
Fig. 6.19

Proof:

In ΔABC, DE || AC (Given)

By Basic Proportionality Theorem: BD/DA = BE/EC ...(1)

In ΔABE, DF || AE (Given)

By Basic Proportionality Theorem: BD/DA = BF/FE ...(2)

From (1) and (2): BE/EC = BF/FE

Hence proved.

5. In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
[Refer to Fig. 6.20 in the textbook]
Fig. 6.20

Proof:

In ΔPOQ, DE || OQ (Given)

By Basic Proportionality Theorem: PD/DO = PE/EQ ...(1)

In ΔPOR, DF || OR (Given)

By Basic Proportionality Theorem: PD/DO = PF/FR ...(2)

From (1) and (2): PE/EQ = PF/FR

By converse of Basic Proportionality Theorem, EF || QR

Hence proved.

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.
[Refer to Fig. 6.21 in the textbook]
Fig. 6.21

Proof:

In ΔOPQ, AB || PQ (Given)

By Basic Proportionality Theorem: OA/AP = OB/BQ ...(1)

In ΔOPR, AC || PR (Given)

By Basic Proportionality Theorem: OA/AP = OC/CR ...(2)

From (1) and (2): OB/BQ = OC/CR

By converse of Basic Proportionality Theorem, BC || QR

Hence proved.

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.

Proof:

Let ABC be a triangle with D as the midpoint of AB.

Draw DE || BC, intersecting AC at E.

By Basic Proportionality Theorem (Theorem 6.1):

AD/DB = AE/EC

Since D is the midpoint of AB, AD = DB

So, AD/DB = 1

Therefore, AE/EC = 1, which means AE = EC

Hence, E is the midpoint of AC, and DE bisects AC.

8. 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.

Proof:

Let ABC be a triangle with D and E as midpoints of AB and AC respectively.

Join DE.

Since D and E are midpoints:

AD = DB and AE = EC

So, AD/DB = 1 and AE/EC = 1

Therefore, AD/DB = AE/EC

By converse of Basic Proportionality Theorem (Theorem 6.2), DE || BC

Hence, the line joining the midpoints of two sides of a triangle is parallel to the third side.

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

Proof:

In trapezium ABCD, AB || DC

Draw EO || AB || DC, where E is on AD

In ΔADC, EO || DC

By Basic Proportionality Theorem: AE/ED = AO/OC ...(1)

In ΔABD, EO || AB

By Basic Proportionality Theorem: AE/ED = BO/OD ...(2)

From (1) and (2): AO/OC = BO/OD

Therefore, AO/BO = CO/DO

Hence proved.

10. 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.

Proof:

Given: AO/BO = CO/DO

This can be rewritten as AO/CO = BO/DO

In ΔAOB and ΔCOD:

AO/CO = BO/DO (Given)

∠AOB = ∠COD (Vertically opposite angles)

Therefore, ΔAOB ~ ΔCOD (SAS similarity criterion)

So, ∠OAB = ∠OCD (Corresponding angles of similar triangles)

But these are alternate interior angles for lines AB and CD with transversal AC

Therefore, AB || CD

Hence, 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:
[Refer to Fig. 6.34 in the textbook]
Fig. 6.34

(i) ΔABC ~ ΔPQR by AAA similarity criterion (all angles equal: 80°, 60°, 40°)

(ii) ΔABC ~ ΔQRP by SSS similarity criterion (sides proportional: AB/QR = BC/RP = CA/PQ)

(iii) Not similar (angles not equal, sides not proportional)

(iv) ΔMNL ~ ΔQPR by SAS similarity criterion (MN/QP = ML/QR and included angle equal)

(v) Not similar (angles not equal)

(vi) ΔDEF ~ ΔPQR by AAA similarity criterion (all angles equal: 80°, 40°, 60°)

2. In Fig. 6.35, ΔODC ~ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
[Refer to Fig. 6.35 in the textbook]
Fig. 6.35

Since ΔODC ~ ΔOBA:

∠DOC = ∠BOA (Corresponding angles)

∠DCO = ∠OAB (Corresponding angles)

∠CDO = ∠OBA = 70° (Corresponding angles)

In quadrilateral ABCD, sum of angles = 360°

∠DOC + ∠BOC = 180° (Linear pair)

∠DOC = 180° - 125° = 55°

In ΔODC, sum of angles = 180°

∠DCO = 180° - (55° + 70°) = 55°

Therefore, ∠OAB = ∠DCO = 55°

So, ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°

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.

Proof:

In trapezium ABCD, AB || DC

In ΔAOB and ΔCOD:

∠OAB = ∠OCD (Alternate interior angles, AB || DC)

∠OBA = ∠ODC (Alternate interior angles, AB || DC)

∠AOB = ∠COD (Vertically opposite angles)

Therefore, ΔAOB ~ ΔCOD (AAA similarity criterion)

So, OA/OC = OB/OD (Corresponding sides of similar triangles)

Hence proved.

4. In Fig. 6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.
[Refer to Fig. 6.36 in the textbook]
Fig. 6.36

Proof:

Given: QR/QS = QT/PR and ∠1 = ∠2

From ∠1 = ∠2, we get PR = PQ (Sides opposite equal angles)

So, QR/QS = QT/PQ

Or, QR/QT = QS/PQ

In ΔPQS and ΔTQR:

QR/QT = QS/PQ (Proved above)

∠PQS = ∠TQR (Common angle)

Therefore, ΔPQS ~ ΔTQR (SAS similarity criterion)

Hence proved.

5. S and T are points on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Proof:

In ΔRPQ and ΔRTS:

∠P = ∠RTS (Given)

∠PRQ = ∠TRS (Common angle)

Therefore, ΔRPQ ~ ΔRTS (AA similarity criterion)

Hence proved.

6. In Fig. 6.37, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.
[Refer to Fig. 6.37 in the textbook]
Fig. 6.37

Proof:

Since ΔABE ≅ ΔACD:

AB = AC and AE = AD (Corresponding parts of congruent triangles)

So, AB/AC = 1 and AD/AE = 1

Therefore, AB/AC = AD/AE

In ΔADE and ΔABC:

AB/AC = AD/AE (Proved above)

∠A = ∠A (Common angle)

Therefore, ΔADE ~ ΔABC (SAS similarity criterion)

Hence proved.

7. In Fig. 6.38, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
[Refer to Fig. 6.38 in the textbook]
Fig. 6.38

(i) ΔAEP ~ ΔCDP

In ΔAEP and ΔCDP:

∠AEP = ∠CDP = 90° (Given, altitudes)

∠APE = ∠CPD (Vertically opposite angles)

Therefore, ΔAEP ~ ΔCDP (AA similarity criterion)

(ii) ΔABD ~ ΔCBE

In ΔABD and ΔCBE:

∠ADB = ∠CEB = 90° (Given, altitudes)

∠ABD = ∠CBE (Common angle)

Therefore, ΔABD ~ ΔCBE (AA similarity criterion)

(iii) ΔAEP ~ ΔADB

In ΔAEP and ΔADB:

∠AEP = ∠ADB = 90° (Given, altitudes)

∠EAP = ∠DAB (Common angle)

Therefore, ΔAEP ~ ΔADB (AA similarity criterion)

(iv) ΔPDC ~ ΔBEC

In ΔPDC and ΔBEC:

∠PDC = ∠BEC = 90° (Given, altitudes)

∠PCD = ∠BCE (Common angle)

Therefore, ΔPDC ~ ΔBEC (AA similarity criterion)

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

Proof:

In parallelogram ABCD, AB || CD and AD || BC

In ΔABE and ΔCFB:

∠A = ∠C (Opposite angles of parallelogram)

∠ABE = ∠CFB (Alternate interior angles, AB || CD)

Therefore, ΔABE ~ ΔCFB (AA similarity criterion)

Hence proved.

9. In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:
[Refer to Fig. 6.39 in the textbook]
Fig. 6.39

(i) ΔABC ~ ΔAMP

In ΔABC and ΔAMP:

∠ABC = ∠AMP = 90° (Given)

∠A = ∠A (Common angle)

Therefore, ΔABC ~ ΔAMP (AA similarity criterion)

(ii) CA/PA = BC/MP

Since ΔABC ~ ΔAMP (Proved above)

CA/PA = BC/MP (Corresponding sides of similar triangles)

Hence proved.

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, show that:

(i) CD/GH = AC/FG

Since ΔABC ~ ΔFEG:

∠A = ∠F, ∠B = ∠E, ∠C = ∠G (Corresponding angles)

Also, AC/FG = BC/EG = AB/EF (Corresponding sides)

Since CD and GH are angle bisectors:

∠ACD = ∠DCB = ½∠C and ∠EGH = ∠HGF = ½∠G

But ∠C = ∠G, so ½∠C = ½∠G, i.e., ∠ACD = ∠EGH

In ΔACD and ΔFGH:

∠A = ∠F (Corresponding angles of similar triangles)

∠ACD = ∠FGH (Proved above)

Therefore, ΔACD ~ ΔFGH (AA similarity criterion)

So, CD/GH = AC/FG (Corresponding sides of similar triangles)

(ii) ΔDCB ~ ΔHGE

In ΔDCB and ΔHGE:

∠B = ∠E (Corresponding angles of similar triangles)

∠DCB = ∠EGH (Both are half of equal angles)

Therefore, ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) ΔDCA ~ ΔHGF

This is the same as part (i), already proved.

11. 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.
[Refer to Fig. 6.40 in the textbook]
Fig. 6.40

Proof:

In isosceles ΔABC, AB = AC, so ∠B = ∠C

In ΔABD and ΔECF:

∠ADB = ∠EFC = 90° (Given, perpendiculars)

∠ABD = ∠ECF (∠B = ∠C, as proved above)

Therefore, ΔABD ~ ΔECF (AA similarity criterion)

Hence proved.

12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig. 6.41). Show that ΔABC ~ ΔPQR.
[Refer to Fig. 6.41 in the textbook]
Fig. 6.41

Proof:

Given: AB/PQ = BC/QR = AD/PM

Since AD and PM are medians, BD = DC and QM = MR

So, BC = 2BD and QR = 2QM

Therefore, AB/PQ = (2BD)/(2QM) = BD/QM

In ΔABD and ΔPQM:

AB/PQ = BD/QM (Proved above)

AD/PM = AB/PQ (Given)

Therefore, ΔABD ~ ΔPQM (SSS similarity criterion)

So, ∠B = ∠Q (Corresponding angles of similar triangles)

In ΔABC and ΔPQR:

AB/PQ = BC/QR (Given)

∠B = ∠Q (Proved above)

Therefore, ΔABC ~ ΔPQR (SAS similarity criterion)

Hence proved.

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

Proof:

In ΔABC and ΔADC:

∠BAC = ∠ADC (Given)

∠ACB = ∠ACD (Common angle)

Therefore, ΔABC ~ ΔDAC (AA similarity criterion)

So, CA/CD = CB/CA (Corresponding sides of similar triangles)

Therefore, 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.

Proof:

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

Extend AD to E such that AD = DE, and extend PM to N such that PM = MN

Join BE, CE, QN, RN

Since AD is median, BD = DC, and AD = DE, so ABEC is a parallelogram

Similarly, PQNR is a parallelogram

In ΔABE and ΔPQN:

AB/PQ = AE/PN = (2AD)/(2PM) = AD/PM

Also, BE = AC and QN = PR (Opposite sides of parallelogram)

So, BE/QN = AC/PR = AB/PQ (Given)

Therefore, ΔABE ~ ΔPQN (SSS similarity criterion)

So, ∠BAE = ∠QPN

Similarly, ∠CAE = ∠RPN

Adding, ∠BAC = ∠QPR

In ΔABC and ΔPQR:

AB/PQ = AC/PR (Given)

∠A = ∠P (Proved above)

Therefore, ΔABC ~ ΔPQR (SAS similarity criterion)

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.

Let AB be the pole and AC be its shadow

Let DE be the tower and DF be its shadow

AB = 6 m, AC = 4 m, DF = 28 m, DE = ?

At the same time, the sun's rays make the same angle with the ground

So, ΔABC ~ ΔDEF (AA similarity criterion)

Therefore, AB/DE = AC/DF

6/DE = 4/28

DE = (6 × 28)/4 = 42 m

Hence, the height of the tower is 42 m.

16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR, prove that AB/PQ = AD/PM.

Proof:

Since ΔABC ~ ΔPQR:

AB/PQ = BC/QR = AC/PR ...(1)

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R

Since AD and PM are medians:

BD = DC and QM = MR

So, BC = 2BD and QR = 2QM

From (1): AB/PQ = (2BD)/(2QM) = BD/QM ...(2)

In ΔABD and ΔPQM:

AB/PQ = BD/QM (From 2)

∠B = ∠Q (Corresponding angles of similar triangles)

Therefore, ΔABD ~ ΔPQM (SAS similarity criterion)

So, AB/PQ = AD/PM (Corresponding sides of similar triangles)

Hence proved.

Theorems & Examples

Theorem 6.1 (Basic Proportionality Theorem or Thales Theorem)

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.

In ΔABC, if DE || BC, then AD/DB = AE/EC

Theorem 6.2 (Converse of Basic Proportionality Theorem)

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

In ΔABC, if AD/DB = AE/EC, then DE || BC

Theorem 6.3 (AAA Similarity Criterion)

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

Theorem 6.4 (SSS Similarity Criterion)

If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar.

Theorem 6.5 (SAS Similarity Criterion)

If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

Theorem 6.6

The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Theorem 6.7 (Pythagoras Theorem)

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Theorem 6.8 (Converse of Pythagoras Theorem)

In a triangle, if the square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Example 1

If a line intersects sides AB and AC of a ΔABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC.

Solution:

In ΔABC, DE || BC (Given)

By Basic Proportionality Theorem: AD/DB = AE/EC

Adding 1 to both sides: (AD/DB) + 1 = (AE/EC) + 1

(AD + DB)/DB = (AE + EC)/EC

AB/DB = AC/EC

Or, DB/AB = EC/AC

Now, AD/AB = (AB - DB)/AB = 1 - DB/AB = 1 - EC/AC = (AC - EC)/AC = AE/AC

Hence, AD/AB = AE/AC

Example 2

ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF || AB. Show that AE/ED = BF/FC.

Solution:

Join AC, intersecting EF at G

In ΔADC, EG || DC (Since EF || AB and AB || DC)

By Basic Proportionality Theorem: AE/ED = AG/GC ...(1)

In ΔABC, GF || AB (Since EF || AB)

By Basic Proportionality Theorem: BF/FC = AG/GC ...(2)

From (1) and (2): AE/ED = BF/FC

Hence proved.

Important Terms

Similar Figures: Two figures having the same shape but not necessarily the same size.

Similar Triangles: Two triangles are similar if their corresponding angles are equal and corresponding sides are proportional.

Ratio of Similarity: The constant ratio of the corresponding sides of two similar triangles.

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