Chapter 7: Triangles (Exercise 7.2)

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 O. Join A to O. Show that:
1. OB = OC
2. AO bisects ∠A.

Solution:
1. To prove OB = OC:
* In ΔABC, given AB = AC ⇒ ∠B = ∠C (Angles opposite to equal sides are equal).
* Halving both sides: ½ ∠B = ½ ∠C ⇒ ∠OBC = ∠OCB.
* In ΔOBC, since ∠OBC = ∠OCB, the sides opposite to equal angles are equal:
OB = OC.

2. To prove AO bisects ∠A:
In ΔAOB and ΔAOC:
* AB = AC (Given)
* OB = OC (Proved above)
* AO = AO (Common side)
* By SSS Congruence Rule, ΔAOB ≅ ΔAOC.
* By CPCT, ∠BAO = ∠CAO.
* Hence, AO bisects ∠A.

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

Solution:
In ΔADB and ΔADC:
* BD = CD (Since AD bisects BC)
* ∠ADB = ∠ADC = 90° (Since AD ⊥ BC)
* AD = AD (Common side)

By SAS Congruence Rule:
ΔADB ≅ ΔADC

By CPCT:
AB = AC
Therefore, ΔABC 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. Show that these altitudes are equal.

Solution:
In ΔABE and ΔACF:
* ∠AEB = ∠AFC = 90° (Given altitudes)
* ∠A = ∠A (Common angle)
* AB = AC (Given isosceles triangle)

By AAS Congruence Rule:
ΔABE ≅ ΔACF

By CPCT:
BE = CF (The altitudes are equal).

Question 4:
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that:
1. ΔABE ≅ ΔACF
2. AB = AC, i.e., ABC is an isosceles triangle.

Solution:
1. In ΔABE and ΔACF:
* ∠AEB = ∠AFC = 90° (Given BE ⊥ AC and CF ⊥ AB)
* ∠A = ∠A (Common angle)
* BE = CF (Given equal altitudes)
By AAS Congruence Rule, ΔABE ≅ ΔACF.

2. To prove AB = AC:
Since ΔABE ≅ ΔACF, by CPCT:
AB = AC.
Hence, ABC is an isosceles triangle.

Question 5:
ΔABC and ΔDBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

Solution:
* In ΔABC (Isosceles with AB = AC):
∠ABC = ∠ACB --- (Equation 1)

* In ΔDBC (Isosceles with DB = DC):
∠DBC = ∠DCB --- (Equation 2)

* Adding Equation 1 and Equation 2:
∠ABC + ∠DBC = ∠ACB + ∠DCB
∠ABD = ∠ACD
(Hence proved.)

Question 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.

Solution:
* In ΔABC, AB = AC ⇒ ∠ABC = ∠ACB = x.
* In ΔADC, AD = AC (since AD = AB = AC) ⇒ ∠ADC = ∠ACD = y.

* In ΔBCD, sum of all interior angles = 180°:
∠B + ∠BCD + ∠D = 180°
x + (x + y) + y = 180°
2x + 2y = 180°
2(x + y) = 180°
x + y = 90°

* Since ∠BCD = x + y, ∠BCD = 90° (Right angle).

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

Solution:
* In ΔABC, given AB = AC ⇒ ∠B = ∠C.
* Sum of angles in a triangle = 180°:
∠A + ∠B + ∠C = 180°
90° + ∠B + ∠B = 180°
2 ∠B = 90°
∠B = 45°

Therefore, ∠B = 45° and ∠C = 45°.

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

Solution:
* Let ABC be an equilateral triangle where AB = BC = AC.
1. Since AB = AC ⇒ ∠B = ∠C.
2. Since BC = AB ⇒ ∠A = ∠C.
Thus, ∠A = ∠B = ∠C.

* Sum of interior angles = 180°:
∠A + ∠B + ∠C = 180°
3 ∠A = 180°
∠A = 60°

Hence, each angle of an equilateral triangle measures 60°.