Exercise 7.2 Solutions (Class 9)
Triangles
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) Show OB = OC:
In ΔABC, AB = AC (Given).
Therefore, ∠ACB = ∠ABC (Angles opposite to equal sides are equal).
(1/2)∠ACB = (1/2)∠ABC.
Since OC and OB are bisectors, this means ∠OCB = ∠OBC.
Now, in ΔOBC, sides opposite to equal angles are equal.
So, OB = OC. (Proved)
(ii) Show AO bisects ∠A:
In ΔABO and ΔACO:
• AB = AC (Given)
• AO = AO (Common side)
• OB = OC (Proved above)
By SSS congruence rule, ΔABO ≅ ΔACO.
Therefore, ∠BAO = ∠CAO (by CPCT).
This means AO bisects ∠A. (Proved)
2. In ΔABC, AD is the perpendicular bisector of BC. Show that ΔABC is an isosceles triangle in which AB = AC.
>Proof:
In ΔADB and ΔADC:
• AD = AD (Common side)
• ∠ADB = ∠ADC = 90° (AD is perpendicular to BC)
• BD = CD (AD bisects BC)
By SAS congruence rule, ΔADB ≅ ΔADC.
Since the triangles are congruent, their corresponding parts are equal.
Therefore, AB = AC (by CPCT).
Hence, ΔABC is an isosceles triangle. (Proved)
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.
>Proof:
In ΔABE and ΔACF:
• ∠A = ∠A (Common angle)
• ∠AEB = ∠AFC = 90° (Given altitudes)
• AB = AC (Given)
By AAS congruence rule, ΔABE ≅ ΔACF.
Since the triangles are congruent, their corresponding parts are equal.
Therefore, BE = CF (by CPCT).
(Proved)
4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that (i) ΔABE ≅ ΔACF (ii) AB = AC, i.e., ABC is an isosceles triangle.
>(i) ΔABE ≅ ΔACF:
In ΔABE and ΔACF:
• ∠A = ∠A (Common angle)
• ∠AEB = ∠AFC = 90° (Given altitudes)
• BE = CF (Given equal altitudes)
By AAS congruence rule, ΔABE ≅ ΔACF. (Proved)
(ii) AB = AC:
Since ΔABE ≅ ΔACF, their corresponding parts are equal.
Therefore, AB = AC (by CPCT).
This means ΔABC is an isosceles triangle. (Proved)
5. ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.
>Proof:
In ΔABC, AB = AC (Given). So, ∠ABC = ∠ACB. —(1)
In ΔDBC, DB = DC (Given). So, ∠DBC = ∠DCB. —(2)
Adding equation (1) and (2):
∠ABC + ∠DBC = ∠ACB + ∠DCB
From the figure, this gives: ∠ABD = ∠ACD.
(Proved)
6. ΔABC 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.
>In ΔABC, AB = AC, so ∠ACB = ∠ABC. —(1)
Given AD = AB, and AB = AC, so AD = AC.
In ΔADC, AD = AC, so ∠ACD = ∠ADC. —(2)
Now consider the large triangle ΔDBC. The sum of its angles is 180°.
∠DBC + ∠BCD + ∠BDC = 180°.
∠ABC + (∠ACB + ∠ACD) + ∠ADC = 180°.
Using (1) and (2), we can substitute:
∠ACB + (∠ACB + ∠ACD) + ∠ACD = 180°.
2∠ACB + 2∠ACD = 180° => 2(∠ACB + ∠ACD) = 180°.
2∠BCD = 180° => ∠BCD = 90°.
(Proved)
7. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
In ΔABC, since AB = AC, it is an isosceles triangle.
Angles opposite to equal sides are equal, so ∠C = ∠B.
By Angle Sum Property for ΔABC:
∠A + ∠B + ∠C = 180°
90° + ∠B + ∠B = 180°
2∠B = 90° => ∠B = 45°.
Since ∠C = ∠B, then ∠C = 45°.
Answer: ∠B = 45° and ∠C = 45°.
8. Show that the angles of an equilateral triangle are 60° each.
Given: ΔABC is equilateral, so AB = BC = CA.
To Prove: ∠A = ∠B = ∠C = 60°.
Proof:
Since AB = BC, then ∠C = ∠A (Angles opposite equal sides). —(1)
Since BC = CA, then ∠A = ∠B (Angles opposite equal sides). —(2)
From (1) and (2), we have ∠A = ∠B = ∠C.
By Angle Sum Property: ∠A + ∠B + ∠C = 180°.
∠A + ∠A + ∠A = 180° => 3∠A = 180° => ∠A = 60°.
Therefore, ∠A = ∠B = ∠C = 60°.
(Proved)
