Chapter 7 Triangles Ex 7.2 Solutions
Question - 1 : - In an isosceles triangle ABC, with AB = AC, the bisectorsof ∠B and ∠C intersect each other at O. Join A to O.Show that:
(i) OB = OC (ii) AO bisects ∠A
Answer - 1 : -
(i) It is given that in triangle ABC, AB = AC
⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of atriangle are equal)
⇒ 
∠ACB = ∠ABC ⇒ ∠OCB = ∠OBC
⇒ OB = OC (Sides opposite to equal angles of a triangle arealso equal)
(ii) In ΔOAB and ΔOAC,
AO =AO (Common)
AB = AC (Given)
OB = OC (Proved above)
Therefore, ΔOAB ≅ ΔOAC (BySSS congruence rule)
⇒ ∠BAO = ∠CAO (CPCT)
⇒ AO bisects ∠A.
Question - 2 : - In ΔABC, AD is the perpendicular bisector of BC (see thegiven figure). Show that ΔABC is an isosceles triangle in which AB = AC.
Answer - 2 : -
In ΔADC and ΔADB,
AD = AD (Common)
∠ADC =∠ADB (Each 90º)
CD = BD (AD is the perpendicular bisector of BC)
∴ ΔADC ≅ ΔADB (By SAS congruence rule)
∴AB = AC (By CPCT)
Therefore, ABC is an isosceles triangle inwhich AB = AC.
Question - 3 : - ABC is an isosceles triangle in which altitudes BE and CFare drawn to equal sides AC and AB respectively (see the given figure). Showthat these altitudes are equal.
Answer - 3 : -
In ΔAEB and ΔAFC,
∠AEB and ∠AFC (Each 90º)
∠A = ∠A (Common angle)
AB = AC (Given)
∴ ΔAEB ≅ ΔAFC (By AAS congruence rule)
⇒ BE = CF (By CPCT)
Question - 4 : - ABC is a triangle in which altitudes BE and CF to sides ACand AB are equal (see the given figure). Show that
(i) 
ABE ≅ ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Answer - 4 : -
(i) In ΔABE and ΔACF,
∠AEB = ∠AFC (Each90º)
∠A = ∠A (Common angle)
BE = CF (Given)
∴ ΔABE ≅ ΔACF (By AAS congruence rule)
(ii) It has already been proved that
ΔABE ≅ ΔACF
⇒ AB = AC (By CPCT)
Question - 5 : - ABC and DBC are two isosceles triangles on the same baseBC (see the given figure). Show that ∠ABD = ∠ACD.
Answer - 5 : -
Let us join AD.
In ΔABD and ΔACD,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common side)
∴ ΔABD 
ΔACD (By SSS congruence rule)
⇒ ∠ABD = ∠ACD (By CPCT)
ΔABC is an isosceles triangle in which AB = AC. Side BA isproduced to D such that AD = AB (see the given figure). Show that ∠BCD is a right angle.
Answer - 6 : -
In ΔABC,
AB = AC (Given)
⇒ ∠ACB = ∠ABC (Angles opposite to equal sides of atriangle are also equal)
In ΔACD,
AC = AD
⇒ ∠ADC = ∠ACD (Angles opposite to equal sides of atriangle are also equal)
In ΔBCD,
∠ABC + ∠BCD + ∠ADC = 180º(Angle sum property of a triangle)
⇒ ∠ACB + ∠ACB +∠ACD + ∠ACD = 180º
⇒ 2(∠ACB + ∠ACD) = 180º
⇒ 2(∠BCD) = 180º
⇒ ∠BCD = 90º
Question - 7 : - ABC is a right angled triangle in which ∠A = 90º and AB = AC. Find ∠B and ∠C.
Answer - 7 : - 
It is given that
AB = AC
⇒ ∠C = ∠B (Angles opposite to equal sides are also equal)
In ΔABC,
∠A + ∠B + ∠C = 180º (Angle sum property of a triangle)
⇒ 90º + ∠B + ∠C = 180º
⇒ 90º + ∠B + ∠B = 180º
⇒ 2 ∠B = 90º
⇒ ∠B = 45º
∴ ∠B = ∠C = 45º
Show that the angles of an equilateraltriangle are 60º each.
Answer - 8 : - 
Let us consider that ABC is an equilateral triangle.
Therefore, AB = BC = AC
AB = AC
⇒ ∠C = ∠B (Angles opposite to equal sides of a triangle are equal)
Also,
AC = BC
⇒ ∠B = ∠A (Angles opposite to equal sides of a triangle are equal)
Therefore, we obtain
∠A = ∠B = ∠C
In ΔABC,
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠A + ∠A = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°
⇒ ∠A = ∠B = ∠C = 60°
Hence, in an equilateral triangle, all interior anglesare of measure 60º