Triangles

Triangles is a chapter in geometry that focuses on the properties and relationships of triangles. It builds upon the concepts introduced in the chapters on lines and angles.

Key Topics

• Congruence of Triangles: Understanding when two triangles are congruent (identical in shape and size).
  • Criteria for congruence: SSS, SAS, ASA, RHS.
• Properties of Triangles:
  • Angle sum property: The angles of a triangle always add up to 180°.
  • Inequality theorem: The sum of any two sides of a triangle is greater than the length of the third side.
  • Exterior angle property: The exterior angle of a triangle is equal to the sum of the two non-adjacent interior angles.
• Similarity of Triangles: The concept of similar triangles (same shape, different size) and their properties.

Importance

The properties and theorems related to triangles are fundamental to many areas of mathematics and science. They are used extensively in geometry, trigonometry, and other branches of mathematics.

Exercise 7.1

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

Ans : 

Given:

• AC = AD
• AB bisects angle A

To prove:

• Triangle ABC is congruent to triangle ABD
• BC = BD

Proof: In triangles ABC and ABD,

• AC = AD (Given)
• Angle CAB = angle DAB 
• (AB bisects angle A)
• AB is common to both triangles.

Therefore, by the SAS (Side-Angle-Side) congruence rule, triangle ABC is congruent to triangle ABD.

Hence, BC = BD (Corresponding Parts of Congruent Triangles are equal, or CPCT).

2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that

(i) ∆ABD ≅ ∆BAC

(ii) BD = AC

(iii) ∠ABD = ∠ BAC

Ans : 

Proof:

i) ∆ABD ≅ ∆BAC

Consider triangles ABD and BAC:

• AD = BC (Given)
• AB is common to both triangles.
• ∠DAB = ∠CBA (Given)

Therefore, by the SAS (Side-Angle-Side) congruence criterion, ∆ABD ≅ ∆BAC.

ii) BD = AC Since ∆ABD ≅ ∆BAC, their corresponding parts are equal. Therefore, BD = AC.

iii) ∠ABD = ∠BAC Again, since ∆ABD ≅ ∆BAC, their corresponding parts are equal. Therefore, ∠ABD = ∠BAC.

Hence, the given statements are proved.

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

Ans : 

In ∆BOC and ∆AOD, we have

∠BOC = ∠AOD

BC = AD [Given]

∠BOC = ∠AOD [Vertically opposite angles]

∴ ∆OBC ≅ ∆OAD [By AAS congruency]

⇒ OB = OA [By C.P.C.T.]

i.e., O is the mid-point of AB.

Thus, CD bisects AB.

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA.

Ans : 

∵ p || q and AC is a transversal,

∴ ∠BAC = ∠DCA …(1)

[Alternate interior angles]

∴ ∠BCA = ∠DAC …(2)

[Alternate interior angles]

Now, in ∆ABC and ∆CDA, we have

∠BAC = ∠DCA [From (1)]

CA = AC [Common]

∠BCA = ∠DAC [From (2)]

∴ ∆ABC ≅ ∆CDA [By ASA congruency]

5. Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that

(i) ∆APB ≅ ∆AQB

(ii) BP = BQ or B is equidistant from the arms ot ∠A.

Ans : 

∴ ∠QAB = ∠PAB

∠Q = ∠P [Each 90°]

∠ABQ = ∠ABP

[By angle sum property of A]

Now, in ∆APB and ∆AQB, we have

∠ABP = ∠ABQ [Proved above]

AB = BA [Common]

∠PAB = ∠QAB [Given]

∴ ∆APB ≅ ∆AQB [By ASA congruency]

Since ∆APB ≅ ∆AQB

⇒ BP = BQ [By C.P.C.T.]

= [Perpendicular distance of B from AQ]

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

Ans : 

Step 1: Analyze the Given Information

We have information about two pairs of equal sides and one pair of equal angles. This suggests using the SAS (Side-Angle-Side) congruence rule.

Step 2: Create Congruent Triangles

To apply SAS congruence, we need to find two triangles with the given information.

• Consider triangles BAC and DAE.
• We are given that AB = AD 
• and AC = AE.
• We also need to prove that angle BAC = angle DAE to apply SAS congruence.

Step 3: Prove Equal Angles

• We know that angle BAD = angle EAC (given).
• Adding angle DAC to both sides, we get:
  • Angle BAD + angle DAC = angle EAC + angle DAC
  • This simplifies to angle BAC = angle DAE.

Step 4: Apply SAS Congruence

Now, we have:

• AB = AD
• Angle BAC = angle DAE
• AC = AE

Therefore, by the SAS congruence rule, triangle BAC is congruent to triangle DAE.

Step 5: Conclusion

Since triangles BAC and DAE are congruent, their corresponding parts are equal. Therefore, BC = DE.

Hence proved.

7. AS 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. (see figure). Show that

(i) ∆DAP ≅ ∆EBP

(ii) AD = BE

Ans : 

We have, P is the mid-point of AB.
∴ AP = BP
∠EPA = ∠DPB [Given]
Adding ∠EPD on both sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ ∠APD = ∠BPE

(i)

∠PAD = ∠PBE [ ∵∠BAD = ∠ABE]
AP = BP [Proved above]
∠DPA = ∠EPB [Proved above]
∴ ∆DAP ≅ ∆EBP [By ASA congruency]

(ii) Since, ∆ DAP ≅ ∆ EBP
⇒ AD = BE [By C.P.C.T.]

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 (see figure). Show that

(i) ∆AMC ≅ ∆BMD

(ii) ∠DBC is a right angle

(iii) ∆DBC ≅ ∆ACB

(iv) CM = 1/2AB

Ans : 

i) ΔAMC ≅ ΔBMD

To prove this, let’s consider the two triangles AMC and BMD.

• AM = BM (Given, M is the midpoint of AB)
• CM = DM (Given)
• ∠AMC = ∠BMD (Vertically opposite angles)

Therefore, by the SAS (Side-Angle-Side) congruence rule, ΔAMC ≅ ΔBMD.

ii) ∠DBC is a right angle 

Since ΔAMC ≅ ΔBMD, their corresponding parts are equal. 

Therefore, ∠ACM = ∠BDM. 

Now, ∠ACM and ∠BDM are alternate interior angles for lines AC and BD with transversal BD.

 Hence, AC is parallel to BD. 

Since ∠ACB is a right angle, and AC is parallel to BD, ∠DBC is also a right angle.

iii) ΔDBC ≅ ΔACB Consider triangles DBC and ACB.

• BC is common to both triangles.
• ∠DBC = ∠ACB (both are right angles)
• BD = AC (from part (i), ΔAMC ≅ ΔBMD)

Therefore, by the RHS (Right Angle – Hypotenuse – Side) congruence rule, ΔDBC ≅ ΔACB.

iv) CM = 1/2 AB Since M is the midpoint of AB, AM = BM = 1/2 AB. From part (i), ΔAMC ≅ ΔBMD, so CM = DM. Therefore, CM = DM = 1/2 AB.

Hence, all four statements are proved.

Exercise 7.2

1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at 0. Join A to 0. Show that

(i) OB = OC

(ii) AO bisects ∠A

Ans : 

Proof:

i) OB = OC

• Since AB = AC (given), 
• Therefore, angle ABC = angle ACB.
• OB bisects angle ABC, so angle OBC = 1/2 * angle ABC.
• OC bisects angle ACB, so angle OCB = 1/2 * angle ACB.
• Since angle ABC = angle ACB, their halves are also equal. Therefore, angle OBC = angle OCB.
• In triangle OBC, angle OBC = angle OCB, which means triangle OBC is an isosceles triangle with OB = OC.

ii) AO bisects ∠A 

To prove that AO bisects angle A, we need to show that angle BAO = angle CAO.

• Consider triangles OAB and OAC.
• AB = AC (given)
• OB = OC (proved in part (i))
• OA is common to both triangles.
• Therefore, by the SSS (Side-Side-Side) congruence rule, triangle OAB is congruent to triangle OAC.
• Since corresponding parts of congruent triangles are equal, angle BAO = angle CAO.

Hence, AO bisects angle A.

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

Ans : 

To Prove: ∆ABC is an isosceles triangle.

Proof:

1. Analyze the given information:
   • AD is perpendicular to BC, so ∠ADB and ∠ADC are right angles.
   • AD bisects BC, so BD = DC.
2. Identify congruent triangles:
   • Consider triangles ABD and ACD.
   • We have:
     • AD is common to both triangles.
     • BD = DC (given)
     • ∠ADB = ∠ADC (both are right angles)
3. Apply congruence criteria:
   • By the Side-Angle-Side (SAS) congruence criterion, triangle ABD is congruent to triangle ACD (AD = AD, ∠ADB = ∠ADC, BD = DC).
4. Conclusion:
   • Since triangles ABD and ACD are congruent, their corresponding parts are equal.
   • Therefore, AB = AC.

Hence, triangle 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 (see figure). Show that these altitudes are equal.

Ans : 

∆ABC is an isosceles triangle.

∴ AB = AC

⇒ ∠ACB = ∠ABC [Angles opposite to equal sides of a A are equal]

⇒ ∠BCE = ∠CBF

Now, in ∆BEC and ∆CFB

∠BCE = ∠CBF [Proved above]

∠BEC = ∠CFB [Each 90°]

BC = CB [Common]

∴ ∆BEC ≅ ∆CFB [By AAS congruency]

So, BE = CF [By C.P.C.T.]

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure).

Show that

(i) ∆ABE ≅ ∆ACF

(ii) AB = AC i.e., ABC is an isosceles triangle.

Ans : 

Proof:

i) ΔABE ≅ ΔACF

To prove the congruence of triangles ABE and ACF, we can use the AAS (Angle-Angle-Side) congruence rule.

• Angle AEB = Angle AFC: Both are right angles as BE and CF are altitudes.
• Angle BAC: This is the common angle to both triangles.
• BE = CF: Given in the problem.

Therefore, by the AAS congruence rule, ΔABE ≅ ΔACF.

ii) AB = AC

Since triangles ABE and ACF are congruent, their corresponding sides are equal.

Therefore, AB = AC.

5. ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ACD.

Ans : 

In ∆ABC, we have

AB = AC [ABC is an isosceles triangle]

∴ ∠ABC = ∠ACB …(1)

[Angles opposite to equal sides of a ∆ are equal]

Again, in ∆BDC, we have

BD = CD [BDC is an isosceles triangle]

∴ ∠CBD = ∠BCD …(2)

[Angles opposite to equal sides of a A are equal]

Adding (1) and (2), we have

∠ABC + ∠CBD = ∠ACB + ∠BCD

⇒ ∠ABD = ∠ACD.

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

Ans : 

AB = AC [Given] …(1)

AB = AD [Given] …(2)

From (1) and (2), we have

AC = AD

Now, in ∆ABC, we have

⇒ 2∠ACB + ∠BAC = 180° …(3)

Similarly, in ∆ACD,

∠ADC + ∠ACD + ∠CAD = 180°

⇒ 2∠ACD + ∠CAD = 180° …(4)

[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]

Adding (3) and (4), we have

2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180°

⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°

⇒ 2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]

⇒ 2∠BCD = 360° – 180° = 180°

⇒ ∠BCD = 180∘2 

    = 90°

Thus, ∠BCD = 90°

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

Ans : 

1. Isosceles Triangle:

Since AB = AC, 

triangle ABC is an isosceles triangle.

2. Angles Opposite Equal Sides:

In an isosceles triangle, the angles opposite to equal sides are equal.

Therefore, angle B = angle C.

3. Angle Sum Property of a Triangle:

So, angle A + angle B + angle C = 180°.

Substituting the given values, we get: 90° + angle B + angle C = 180°.

4. Finding the Angles:

Since angle B = angle C, 

we can substitute angle B for angle C in the equation.

90° + 2 * angle B = 180°.

2 * angle B = 90°.

Angle B = 45°.

Since angle B = angle C, angle C = 45°.

Therefore, angle B = 45° and angle C = 45°.

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

Ans : 

Proof:

• Definition of an Equilateral Triangle: In an equilateral triangle, all sides are equal.
• Let’s denote the angles: Let the angles of the equilateral triangle ABC be ∠A, ∠B, and ∠C.
• Angles opposite equal sides: In a triangle, angles opposite equal sides are equal. Therefore, ∠A = ∠B and ∠B = ∠C.
• So, ∠A + ∠B + ∠C = 180°.
• Substituting equal angles: Since ∠A = ∠B = ∠C, we can substitute any of these with x.
  • Therefore, x + x + x = 180°.
  • This simplifies to 3x = 180°.
  • Hence, x = 60°.

Conclusion: Since ∠A = ∠B = ∠C = x and x = 60°, each angle of the equilateral triangle is 60 degrees.

Therefore, the angles of an equilateral triangle are 60° each.

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 (see figure). 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.

Ans : 

(i) ΔABD ≅ ΔACD

• Given: Triangles ABC and DBC are isosceles with AB = AC and BD = DC.
• Proof:
  • AB = AC (given)
  • BD = DC (given)
  • AD is common to both triangles.
  • Therefore, by the Side-Side-Side (SSS) congruence rule, ΔABD ≅ ΔACD.

(ii) ΔABP ≅ ΔACP

• Proof:
  • Since ΔABD ≅ ΔACD (proved in part (i)), we have angle BAD = angle CAD.
  • AB = AC (given)
  • AP is common to both triangles.
  • Therefore, by the Side-Angle-Side (SAS) congruence rule, ΔABP ≅ ΔACP.

(iii)

• Proof:
  • Since ΔABP ≅ ΔACP, angle BAP = angle CAP. Hence, AP bisects angle A.
  • Similarly, since ΔABD ≅ ΔACD, angle BAD = angle CAD. Hence, AP bisects angle D.

(iv) AP is the perpendicular bisector of BC

• Proof:
  • Since ΔABP ≅ ΔACP, BP = CP. Hence, AP bisects BC.
  • Also, since ΔABD ≅ ΔACD, angle APD = angle CPD.
  • In triangles APB and APC, AP is common, BP = CP, and angle APB = angle APC.
  • Therefore, by the Side-Angle-Side (SAS) congruence rule, ΔAPB ≅ ΔAPC.
  • Hence, angle BPA = angle CPA.
  • Since angle BPA and angle CPA are linear pairs and equal, both must be right angles.
  • Therefore, AP is perpendicular to 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

Ans :

i) AD bisects BC

To prove that AD bisects BC, we need to show that BD = CD.

Consider triangles ABD and ACD:

• AB = AC (given)
• Angle ADB = Angle ADC (both are right angles as AD is an altitude)
• AD is common to both triangles.

Therefore, by the Right Angle-Hypotenuse-Side (RHS) congruence rule, triangle ABD is congruent to triangle ACD.

Hence, BD = CD.

Therefore, AD bisects BC.

ii) AD bisects ∠A

Since triangles ABD and ACD are congruent (proved above), their corresponding angles are equal.

Therefore, angle BAD = angle CAD.

Hence, AD bisects angle A.

3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that

(i) ∆ABC ≅ ∆PQR

(ii) ∆ABM ≅ ∆PQN

Ans : 

• AB = PQ
• BC = QR
• AM is the median of triangle ABC
• PN is the median of triangle PQR

We need to prove:

• Triangle ABM is congruent to triangle PQN
• Triangle ABC is congruent to triangle PQR

Proof:

i) ΔABM ≅ ΔPQN

Given:

• AB = PQ
• AM = PN (as they are medians, they divide the respective sides into equal parts)
• BM = QN (as they are half of equal sides BC and QR)

Therefore, by the Side-Side-Side (SSS) congruence rule, ΔABM ≅ ΔPQN.

ii) ΔABC ≅ ΔPQR

From (i), we know that ΔABM ≅ ΔPQN. This implies that angle ABM = angle PQN.

Now, consider triangles ABC and PQR:

• AB = PQ (given)
• Angle ABM = Angle PQN (proved above)
• BM = QN (given)

Therefore, by the SAS (Side-Angle-Side) congruence rule, ΔABC ≅ ΔPQR.

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

Ans : 

Given:

• BE and CF are altitudes of triangle ABC and BE = CF.

To Prove:

• Triangle ABC is isosceles.

Proof:

1. Identify Right-Angled Triangles:

• Since BE and CF are altitudes, triangles BEC and BFC are right-angled triangles with right angles at E and F, respectively.

2. Apply RHS Congruence:

• In triangles BEC and BFC:
  • BC is common to both triangles.
  • BE = CF (given)
  • ∠BEC = ∠BFC = 90° (right angles)
• Therefore, by the RHS (Right Angle – Hypotenuse – Side) congruence rule, triangle BEC is congruent to triangle BFC.

3. Conclusion:

• Since triangles BEC and BFC are congruent, their corresponding parts are equal.
• Therefore, AB = AC.

Hence, triangle ABC is isosceles.

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

Ans :

∠APB = 90° and ∠APC = 90°

In ∆ABP and ∆ACP, we have

∠APB = ∠APC [Each 90°]

AB = AC [Given]

AP = AP [Common]

∴ ∆ABP ≅ ∆ACP [By RHS congruency]

So, ∠B = ∠C [By C.P.C.T.]