**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 = ****1/2****AB**

**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:**

**Analyze the given information:**- AD is perpendicular to BC, so ∠ADB and ∠ADC are right angles.
- AD bisects BC, so BD = DC.

**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)

**Apply congruence criteria:**- By the Side-Angle-Side (SAS) congruence criterion, triangle ABD is congruent to triangle ACD (AD = AD, ∠ADB = ∠ADC, BD = DC).

**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:**

**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.

**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.

**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.]