NCERT Solutions Class 9th Maths Chapter – 7 Triangles
| Textbook | NCERT |
| Class | 9th |
| Subject | Mathematics |
| Chapter | 7th |
| Chapter Name | Triangles |
| Category | Class 9th Math Solutions |
| Medium | English |
| Source | Last Doubt |
NCERT Solutions Class 9th Maths Chapter – 7 Triangles
Chapter – 7
Triangles
Exercise 7.2
| Question 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 Solution i) in ∆ABC, we have AB = AC [Given] ∴ ∠ABC = ∠ACB [Angles opposite to equal sides of a A are equal]  ⇒ 1/2∠ABC = 1/2∠ACB or ∠OBC = ∠OCB ⇒ OC = OB [Sides opposite to equal angles of a ∆ are equal] (ii) In ∆ABO and ∆ACO, we have
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Question 2. In ∆ABC, AD is the perpendicular bisector of BC (see figure). Show that ∆ ABC is an isosceles triangle in which AB = AC. Solution Since AD is bisector of BC. ∴ BD = CD Now, in ∆ABD and ∆ACD, we have AD = DA [Common] ∠ADB = ∠ADC [Each 90°] BD = CD [Proved above] ∴ ∆ABD ≅ ∆ACD [By SAS congruency] ⇒ AB = AC [By C.P.C.T.] Thus, ∆ABC is an isosceles triangle. |
Question 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. Solution ∆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.] |
| Question 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. Solution (ii) Since, ∆ABE ≅ ∆ACF |
Question 5. ABC and DBC are isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ACD. Solution 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. |
Question 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. Solution AB = AC [Given] …(1) AB = AD [Given] …(2) From (1) and (2), we have AC = AD Now, in ∆ABC, we have ∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A] ⇒ 2∠ACB + ∠BAC = 180° …(3) [∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)] 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° |
| Question 7. ABC is a right angled triangle in which ∠A = 90° and AB = AC, find ∠B and ∠C. Solution In ∆ABC, we have AB = AC [Given] ∴ Their opposite angles are equal.  ⇒ ∠ACB = ∠ABC Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆] ⇒ 90° + ∠B + ∠C = 180° [∠A = 90°(Given)] ⇒ ∠B + ∠C= 180°- 90° = 90° But ∠B = ∠C ∠B = ∠C = 90∘/2 = 45° Thus, ∠B = 45° and ∠C = 45° |
| Question 8. Show that the angles of an equilateral triangle are 60° each. Solution In ∆ABC, we have  AB = BC = CA [ABC is an equilateral triangle] AB = BC ⇒ ∠A = ∠C …(1) [Angles opposite to equal sides of a A are equal] Similarly, AC = BC ⇒ ∠A = ∠B …(2) From (1) and (2), we have ∠A = ∠B = ∠C = x (say) Since, ∠A + ∠B + ∠C = 180° [Angle sum property of a A] ∴ x + x + x = 180o ⇒ 3x = 180° ⇒ x = 60° ∴ ∠A = ∠B = ∠C = 60° Thus, the angles of an equilateral triangle are 60° each. |
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