Chapter 7. Triangles
Exercise 7.2
Q1. 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
Solution:
Given: In an isosceles triangle ABC, with AB = AC, the bisectors of ∠ B and ∠ C intersect each other at O.
To prove:
(i) OB = OC
(ii) AO bisects ∠ A
Proof: In ΔABC, We have:
AB = AC
∠ B = ∠ C [ opposite angle to the equal side ]
Or 1/2 ∠ B = 1/2 ∠C
So, ∠OBC = ∠OCB […1]
In D ABO and D ACO
AB = AC [given]
∠OBC = ∠OCB [from …1]
AO = AO [common]
By SAS Congruence Criterion Rule
DABO ≅ DACO
OB = OC [ By CPCT ]
∠BAO = ∠CAO [ By CPCT ]
AO bisect ∠A.
Q2. In Δ ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that Δ ABC is an isosceles triangle in which AB = AC.
Solution:
Given: In Δ ABC, AD is the perpendicular bisector of BC.
To prove: Δ ABC is an isosceles triangle in which AB = AC.
Proof: In Δ ABD and Δ ACD, we have
DB = DC [since D bisect BC]
∠ BDC = ∠ADC [AD is the perpendicular bisector of BC].
AD = AD [common]
By SAS Congruence Criterion Rule
Δ ABD ≅ Δ ACD
AB =AC [CPCT]
Hence, Δ ABC is an isosceles triangle
Q3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.
Solution:
Given: ABC is an isosceles triangle in which BE ⊥ AC and CF ⊥ AB where AB = AC.
To prove: BE = CF.
Proof: Here, BE ⊥ AC and CF ⊥ AB (Given)
In ΔABE and Δ ACF
∠ AEB = ∠ AFC (90० Each)
∠ A = ∠ A (Common)
AB = AC (Given)
By ASA Congruency Criterion Rule
ΔABE ≅ Δ ACF
BE = CF [ By CPCT ]
Q4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
(i) Δ ABE ≅ Δ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Solution:
Given: ABC is a triangle in which
BE ⊥ AC and CF ⊥ AB and BE = CF
To Prove :
(i) Δ ABE ≅ Δ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Proof :
(i) In Δ ABE and Δ ACF
BE = CF (Given)
∠ AEB = ∠ AFC (90० Each)
∠ A = ∠ A (Common)
Using ASA Congruency Property
Δ ABE ≅ Δ ACF Proved I
(ii) AB = AC [By CPCT]
Therefore, ABC is an isosceles triangle.
Q5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ ABD = ∠ ACD.
Solution:
Given : ABC and DBC are two isosceles triangles on the same base BC.
To prove : ∠ ABD = ∠ ACD
Proof : ABC is an isosceles triangle in which
AB = AC (Given)
∴ ∠ ABC = ∠ ACB .......... (1)
(Opposite angles of equal sides)
Similarily,
BCD is also an isosceles triangle.
BD = CD (Given)
∴ ∠ DBC = ∠ DCB .......... (2)
(Opposite angles of equal sides)
Adding Equation (1) and (2)
∠ ABC + ∠ DBC = ∠ ACB + ∠ DCB
Or, ∠ ABD = ∠ ACD Proved
Q6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig. 7.34). Show that ∠ BCD is a right angle.
Solution :
Given: ΔABC is an isosceles triangle in which AB = AC.
Side BA is produced to D such that AD = AB
To prove : ∠ BCD = 90०
Proof :
AB = AC .............. (1) (Given)
And, AB = AD .............. (2) (Given) 
From equation (1) and (2) we have
AC = AD ...............(3)
∴ ∠3 = ∠4 .... (4) (Opposite angles of equal sides)
Now, AB = AC from (1)
∴ ∠1 = ∠2 .... (5) (Opposite angles of equal sides)
In ΔABC,
Exterior ∠5 = ∠1 + ∠2
Or, ∠5 = ∠2 + ∠2 from (5)
Or, ∠5 = 2∠2 ....... (6)
Similarily,
Exterior ∠6 = ∠3 + ∠4
Or, ∠6 = 2∠3 from (7)
Adding equation (6) and (7)
∠5 + ∠6 = 2∠2 + 2∠3
∠5 + ∠6 = 2(∠2 + ∠3)
Or, 180० = 2(∠2 + ∠3) [ ∵ ∠BAC + ∠DAC = 180० ]
Or, ∠2 + ∠3 = 180० / 2
Or, ∠BCD = 90० Proved
Q7. ABC is a right angled triangle in which ∠ A = 90° and AB = AC. Find ∠ B and ∠ C.
Solution:
Given : ABC is a right angled triangle in which
∠ A = 90° and AB = AC.
To Find : ∠B and ∠C
AB = AC (Given)
∴ ∠B = ∠C ............(1)
(Opposite angles of equal sides)
In triangle ABC,
∠A + ∠B + ∠C = 180० (Angle sum property)
90° + ∠B + ∠B = 180० Using equation (1)
2 ∠B = 180० - 90°
2 ∠B = 90°
∠B = 90°/ 2
∠B = 45°
∴ ∠B = 45° and ∠C = 45°
Q8. Show that the angles of an equilateral triangle are 60° each. ⊥
Solution:
Given: ABC is a equilateral triangle, in which
AB = BC = AC
To Prove :
∠A = ∠B = ∠C = 60°
Proof :
AB = AC (Given)
∠B = ∠C ....................... (1) [opposite angle of equal sides]
AB = BC (Given)
∠A = ∠C ....................... (2) [opposite angle of equal sides]
AC = BC (Given)
∠A = ∠B ....................... (3) [opposite angle of equal sides]
From equation (1), (2) and (3) we have
∠A = ∠B = ∠C .............. (4)
In triangle ABC
∠A + ∠B + ∠C = 180°
∠A + ∠A + ∠A = 180°
3 ∠A = 180°
∠A = 180°/3
∠A = 60°
∴ ∠A = ∠B = ∠C = 60°
