Triangles - NCERT Solutions Chapter 7

Exercise 7.1
Exercise 7.2
Exercise 7.3
Key Concepts

Exercise 7.1

1. In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

In ∆ABC and ∆ABD:

AC = AD (Given)

∠CAB = ∠DAB (AB bisects ∠A)

AB = AB (Common)

Therefore, ∆ABC ≅ ∆ABD (by SAS congruence rule)

Since the triangles are congruent, BC = BD (CPCT)

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that (i) ∆ABD ≅ ∆BAC (ii) BD = AC (iii) ∠ABD = ∠BAC.

(i) In ∆ABD and ∆BAC:

AD = BC (Given)

∠DAB = ∠CBA (Given)

AB = BA (Common)

Therefore, ∆ABD ≅ ∆BAC (by SAS congruence rule)

(ii) Since ∆ABD ≅ ∆BAC, BD = AC (CPCT)

(iii) Since ∆ABD ≅ ∆BAC, ∠ABD = ∠BAC (CPCT)

3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Let CD intersect AB at O.

In ∆AOD and ∆BOC:

AD = BC (Given)

∠DAO = ∠CBO = 90° (Given - perpendiculars)

∠AOD = ∠BOC (Vertically opposite angles)

Therefore, ∆AOD ≅ ∆BOC (by AAS congruence rule)

So, AO = BO (CPCT)

Hence, CD bisects AB at O.

4. l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

In ∆ABC and ∆CDA:

∠BAC = ∠DCA (Alternate angles, as l || m and AC is transversal)

AC = CA (Common)

∠BCA = ∠DAC (Alternate angles, as p || q and AC is transversal)

Therefore, ∆ABC ≅ ∆CDA (by ASA congruence rule)

5. Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A. Show that: (i) ∆APB ≅ ∆AQB (ii) BP = BQ or B is equidistant from the arms of ∠A.

(i) In ∆APB and ∆AQB:

∠PAB = ∠QAB (l is the bisector of ∠A)

AB = AB (Common)

∠APB = ∠AQB = 90° (BP and BQ are perpendiculars)

Therefore, ∆APB ≅ ∆AQB (by AAS congruence rule)

(ii) Since ∆APB ≅ ∆AQB, BP = BQ (CPCT)

Hence, B is equidistant from the arms of ∠A.

6. In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Given: ∠BAD = ∠EAC

Adding ∠DAC to both sides:

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE

In ∆ABC and ∆ADE:

AB = AD (Given)

∠BAC = ∠DAE (Proved above)

AC = AE (Given)

Therefore, ∆ABC ≅ ∆ADE (by SAS congruence rule)

So, BC = DE (CPCT)

7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that (i) ∆DAP ≅ ∆EBP (ii) AD = BE

(i) Given: ∠EPA = ∠DPB

Adding ∠EPD to both sides:

∠EPA + ∠EPD = ∠DPB + ∠EPD

∠DPA = ∠EPB

In ∆DAP and ∆EBP:

∠BAD = ∠ABE (Given)

AP = BP (P is mid-point of AB)

∠DPA = ∠EPB (Proved above)

Therefore, ∆DAP ≅ ∆EBP (by ASA congruence rule)

(ii) Since ∆DAP ≅ ∆EBP, AD = BE (CPCT)

8. In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B. Show that: (i) ∆AMC ≅ ∆BMD (ii) ∠DBC is a right angle (iii) ∆DBC ≅ ∆ACB (iv) CM = ½AB

(i) In ∆AMC and ∆BMD:

AM = BM (M is mid-point of AB)

CM = DM (Given)

∠AMC = ∠BMD (Vertically opposite angles)

Therefore, ∆AMC ≅ ∆BMD (by SAS congruence rule)

(ii) Since ∆AMC ≅ ∆BMD:

AC = BD (CPCT)

∠MAC = ∠MBD (CPCT)

But ∠MAC and ∠MBC are alternate angles

So AC || BD

Since ∠ACB = 90° and AC || BD, ∠DBC = 90° (corresponding angles)

(iii) In ∆DBC and ∆ACB:

BC = CB (Common)

∠DBC = ∠ACB = 90° (Proved above)

DB = AC (Proved in part ii)

Therefore, ∆DBC ≅ ∆ACB (by SAS congruence rule)

(iv) Since ∆DBC ≅ ∆ACB, DC = AB (CPCT)

But DC = DM + MC = CM + CM = 2CM

So, 2CM = AB

Therefore, CM = ½AB

Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that: (i) OB = OC (ii) AO bisects ∠A

(i) Since AB = AC (Given), ∆ABC is isosceles

Therefore, ∠ABC = ∠ACB (Angles opposite to equal sides)

Since BO and CO are bisectors:

∠ABO = ½∠ABC and ∠ACO = ½∠ACB

So, ∠ABO = ∠ACO

In ∆OBC:

∠OBC = ½∠ABC and ∠OCB = ½∠ACB

Since ∠ABC = ∠ACB, we have ∠OBC = ∠OCB

Therefore, OB = OC (Sides opposite to equal angles)

(ii) In ∆ABO and ∆ACO:

AB = AC (Given)

∠ABO = ∠ACO (Proved above)

OB = OC (Proved in part i)

Therefore, ∆ABO ≅ ∆ACO (by SAS congruence rule)

So, ∠BAO = ∠CAO (CPCT)

Hence, AO bisects ∠A

2. In ∆ABC, AD is the perpendicular bisector of BC. Show that ∆ABC is an isosceles triangle in which AB = AC.

Since AD is the perpendicular bisector of BC:

BD = DC and ∠ADB = ∠ADC = 90°

In ∆ABD and ∆ACD:

BD = DC (AD is bisector of BC)

∠ADB = ∠ADC = 90° (AD is perpendicular to BC)

AD = AD (Common)

Therefore, ∆ABD ≅ ∆ACD (by SAS congruence rule)

So, AB = AC (CPCT)

Hence, ∆ABC is an isosceles triangle.

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

In ∆ABE and ∆ACF:

∠AEB = ∠AFC = 90° (BE and CF are altitudes)

∠A = ∠A (Common)

AB = AC (Given - isosceles triangle)

Therefore, ∆ABE ≅ ∆ACF (by AAS congruence rule)

So, BE = CF (CPCT)

Hence, the altitudes are equal.

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that (i) ∆ABE ≅ ∆ACF (ii) AB = AC, i.e., ABC is an isosceles triangle.

(i) In ∆ABE and ∆ACF:

∠AEB = ∠AFC = 90° (BE and CF are altitudes)

∠A = ∠A (Common)

BE = CF (Given)

Therefore, ∆ABE ≅ ∆ACF (by AAS congruence rule)

(ii) Since ∆ABE ≅ ∆ACF, AB = AC (CPCT)

Hence, ABC is an isosceles triangle.

5. ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

Since ∆ABC is isosceles with AB = AC:

∠ABC = ∠ACB ... (1)

Since ∆DBC is isosceles with DB = DC:

∠DBC = ∠DCB ... (2)

Subtracting equation (2) from equation (1):

∠ABC - ∠DBC = ∠ACB - ∠DCB

∠ABD = ∠ACD

6. ∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

Given: AB = AC and AD = AB

Therefore, AD = AB = AC

Since AB = AC, ∠ABC = ∠ACB (Angles opposite to equal sides)

Let ∠ABC = ∠ACB = x

In ∆ACD, since AD = AC, ∆ACD is isosceles

Therefore, ∠ADC = ∠ACD (Angles opposite to equal sides)

Let ∠ADC = ∠ACD = y

In ∆ABC: ∠BAC + ∠ABC + ∠ACB = 180°

∠BAC + x + x = 180°

∠BAC = 180° - 2x ... (1)

In ∆ACD: ∠CAD + ∠ADC + ∠ACD = 180°

∠CAD + y + y = 180°

∠CAD = 180° - 2y ... (2)

Since points D, A, B are collinear (BA is produced to D):

∠BAC + ∠CAD = 180° (linear pair)

(180° - 2x) + (180° - 2y) = 180°

360° - 2x - 2y = 180°

2x + 2y = 180°

x + y = 90° ... (3)

Now, ∠BCD = ∠ACB + ∠ACD = x + y = 90°

Therefore, ∠BCD is a right angle.

7. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Since AB = AC, ∆ABC is isosceles

Therefore, ∠B = ∠C (Angles opposite to equal sides)

In ∆ABC: ∠A + ∠B + ∠C = 180°

90° + ∠B + ∠C = 180°

∠B + ∠C = 90°

Since ∠B = ∠C:

2∠B = 90°

∠B = 45°

Therefore, ∠B = ∠C = 45°

8. Show that the angles of an equilateral triangle are 60° each.

Let ∆ABC be an equilateral triangle with AB = BC = CA

Since AB = AC, ∠B = ∠C (Angles opposite to equal sides)

Since BC = AC, ∠A = ∠B (Angles opposite to equal sides)

Therefore, ∠A = ∠B = ∠C

In ∆ABC: ∠A + ∠B + ∠C = 180°

3∠A = 180°

∠A = 60°

Hence, ∠A = ∠B = ∠C = 60°

Exercise 7.3

1. ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. If AD is extended to intersect BC at P, show that (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP (iii) AP bisects ∠A as well as ∠D (iv) AP is the perpendicular bisector of BC.

(i) In ∆ABD and ∆ACD:

AB = AC (∆ABC is isosceles)

BD = CD (∆DBC is isosceles)

AD = AD (Common)

Therefore, ∆ABD ≅ ∆ACD (by SSS congruence rule)

(ii) In ∆ABP and ∆ACP:

AB = AC (Given)

∠BAP = ∠CAP (From congruent triangles in part i, CPCT)

AP = AP (Common)

Therefore, ∆ABP ≅ ∆ACP (by SAS congruence rule)

(iii) From part (i): ∠BAD = ∠CAD (CPCT), so AP bisects ∠A

Also, ∠BDA = ∠CDA (CPCT), so AP bisects ∠D

(iv) From part (ii): BP = CP (CPCT) and ∠APB = ∠APC (CPCT)

Since ∠APB + ∠APC = 180° (linear pair)

We have ∠APB = ∠APC = 90°

Therefore, AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects ∠A.

(i) In ∆ABD and ∆ACD:

AB = AC (Given)

∠ADB = ∠ADC = 90° (AD is altitude)

AD = AD (Common)

Therefore, ∆ABD ≅ ∆ACD (by RHS congruence rule)

So, BD = CD (CPCT)

Hence, AD bisects BC.

(ii) Since ∆ABD ≅ ∆ACD, ∠BAD = ∠CAD (CPCT)

Hence, AD bisects ∠A.

3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ∆PQR. Show that: (i) ∆ABM ≅ ∆PQN (ii) ∆ABC ≅ ∆PQR

(i) Since AM is median of ∆ABC, BM = ½BC

Since PN is median of ∆PQR, QN = ½QR

Given BC = QR, so BM = QN

In ∆ABM and ∆PQN:

AB = PQ (Given)

BM = QN (Proved above)

AM = PN (Given)

Therefore, ∆ABM ≅ ∆PQN (by SSS congruence rule)

(ii) Since ∆ABM ≅ ∆PQN, ∠ABM = ∠PQN (CPCT)

That is, ∠ABC = ∠PQR

In ∆ABC and ∆PQR:

AB = PQ (Given)

∠ABC = ∠PQR (Proved above)

BC = QR (Given)

Therefore, ∆ABC ≅ ∆PQR (by SAS congruence rule)

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

In ∆BEC and ∆CFB:

∠BEC = ∠CFB = 90° (BE and CF are altitudes)

BC = CB (Common - hypotenuse)

BE = CF (Given)

Therefore, ∆BEC ≅ ∆CFB (by RHS congruence rule)

So, ∠BCE = ∠CBF (CPCT)

That is, ∠ACB = ∠ABC

Therefore, AB = AC (Sides opposite to equal angles)

Hence, ∆ABC is isosceles.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

In ∆APB and ∆APC:

AB = AC (Given)

∠APB = ∠APC = 90° (AP ⊥ BC)

AP = AP (Common)

Therefore, ∆APB ≅ ∆APC (by RHS congruence rule)

So, ∠B = ∠C (CPCT)

Key Concepts and Theorems

SAS Congruence Rule (Axiom 7.1)

Two triangles are congruent if two sides and the included angle of one triangle are equal to two sides and the included angle of the other triangle.

ASA Congruence Rule (Theorem 7.1)

Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.

AAS Congruence Rule

Two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal.

Theorem 7.2

Angles opposite to equal sides of an isosceles triangle are equal.

Theorem 7.3

The sides opposite to equal angles of a triangle are equal (Converse of Theorem 7.2).

SSS Congruence Rule (Theorem 7.4)

If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

RHS Congruence Rule (Theorem 7.5)

If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

Note: RHS stands for Right angle - Hypotenuse - Side

Important Terms

  • Congruent Figures: Figures that are equal in all respects (same shape and same size)
  • CPCT: Corresponding Parts of Congruent Triangles (are equal)
  • Isosceles Triangle: A triangle in which two sides are equal
  • Equilateral Triangle: A triangle in which all three sides are equal
  • Altitude: A perpendicular drawn from a vertex to the opposite side
  • Median: A line segment joining a vertex to the mid-point of the opposite side
  • Hypotenuse: The longest side of a right triangle (opposite to the right angle)

Summary Points

  1. Two figures are congruent if they are of the same shape and of the same size.
  2. Two circles of the same radii are congruent.
  3. Two squares of the same sides are congruent.
  4. If two triangles ABC and PQR are congruent under the correspondence A ↔ P, B ↔ Q and C ↔ R, then symbolically it is expressed as ∆ABC ≅ ∆PQR.
  5. Each angle of an equilateral triangle is of 60°.
  6. In congruent triangles, corresponding parts are always equal.
  7. For triangles to be congruent, it is important that the equal angles are included between the pairs of equal sides (in SAS rule).
  8. SSA or ASS rule does not guarantee congruence.
  9. Equality of three angles is not sufficient for congruence of triangles.
  10. For congruence of triangles, out of three equal parts, one has to be a side.
Important Examples

Example 1: In a triangle, if two sides are equal, prove that the angles opposite to them are also equal.

Solution: This is Theorem 7.2. Draw the angle bisector from the vertex where equal sides meet. Use SAS congruence to prove the triangles formed are congruent, hence opposite angles are equal.

Example 2: AB is a line segment and line l is its perpendicular bisector. If a point P lies on l, show that P is equidistant from A and B.

Solution: Let l intersect AB at C (mid-point). In ∆PCA and ∆PCB, we have AC = BC, ∠PCA = ∠PCB = 90°, and PC = PC (common). So ∆PCA ≅ ∆PCB by SAS rule. Therefore PA = PB (CPCT).

Example 3: P is a point equidistant from two lines l and m intersecting at point A. Show that the line AP bisects the angle between them.

Solution: Let PB ⊥ l and PC ⊥ m with PB = PC. In ∆PAB and ∆PAC, we have PB = PC (given), ∠PBA = ∠PCA = 90°, and PA = PA (common). So ∆PAB ≅ ∆PAC by RHS rule. Therefore ∠PAB = ∠PAC (CPCT), hence AP bisects the angle.

Tips for Solving Problems

  • Identify which triangles need to be proved congruent
  • List out the given equal parts (sides or angles)
  • Look for common sides or angles
  • Identify any pairs of vertically opposite angles or alternate angles
  • Choose the appropriate congruence rule (SAS, ASA, AAS, SSS, or RHS)
  • Use CPCT to conclude about other equal parts
  • Remember: The order of vertices in congruent triangles matters

Common Mistakes to Avoid

  • Not writing the correspondence of vertices correctly when stating congruence
  • Trying to use SSA or ASS rule (these don't work!)
  • Assuming triangles are congruent based on three equal angles only
  • Forgetting to state which congruence rule is being used
  • Not identifying the included angle correctly in SAS rule
  • Not identifying the included side correctly in ASA rule
  • Confusing altitude with median or angle bisector