Class 9 Mathematics
Chapter 7: Triangles
Exercise 7.2 – Stepwise Solutions
Important Results Used
1. In an isosceles triangle, angles opposite equal sides are equal.
2. If two angles of a triangle are equal, then the sides opposite to them are equal.
3. In right triangles, RHS congruence rule can be used.
4. Sum of angles of a triangle is 180°.
5. CPCT: Corresponding parts of congruent triangles are equal.
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
Given: AB = AC, BO and CO are bisectors of ∠B and ∠C respectively.
Since AB = AC, in ΔABC,
∠B = ∠C
Because BO and CO bisect these angles,
∠OBC = 1/2 ∠B and ∠OCB = 1/2 ∠C
But ∠B = ∠C, so
∠OBC = ∠OCB
In ΔBOC, angles at B and C are equal.
Therefore, sides opposite to these equal angles are equal.
Hence, OB = OC
Now consider ΔABO and ΔACO:
AB = AC
OB = OC
AO = AO (common)
Therefore,
ΔABO ≅ ΔACO by SSS.
So, by CPCT,
∠BAO = ∠CAO
Thus, AO bisects ∠A.
Q2. In ΔABC, AD is the perpendicular bisector of BC. Show that ΔABC is an isosceles triangle in which AB = AC.
Given: AD is the perpendicular bisector of BC.
So,
BD = DC
and ∠ADB = ∠ADC = 90°
Now in right triangles ΔADB and ΔADC:
AD = AD (common)
BD = DC
∠ADB = ∠ADC
Therefore,
ΔADB ≅ ΔADC by SAS.
So, by CPCT,
AB = AC
Hence, ΔABC 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. Show that these altitudes are equal.
Given: AB = AC, BE ⟂ AC and CF ⟂ AB
Area of ΔABC using base AB:
Area = 1/2 × AB × CF
Area of ΔABC using base AC:
Area = 1/2 × AC × BE
Since both are areas of the same triangle,
1/2 × AB × CF = 1/2 × AC × BE
But AB = AC
Therefore,
CF = BE
Hence, the altitudes are equal.
Q4. 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.
Given: BE = CF, BE ⟂ AC and CF ⟂ AB
So,
∠AEB = ∠AFC = 90°
Also, ∠BAE = ∠CAF = ∠A
Now in triangles ΔABE and ΔACF:
∠AEB = ∠AFC
∠BAE = ∠CAF
BE = CF
Therefore,
ΔABE ≅ ΔACF by AAS.
Hence, by CPCT,
AB = AC
So, ΔABC is an isosceles triangle.
Q5. ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.
Given:
In ΔABC, AB = AC
In ΔDBC, DB = DC
Since AB = AC, base angles of ΔABC are equal:
∠ABC = ∠ACB
Since DB = DC, base angles of ΔDBC are equal:
∠CBD = ∠BCD
Now,
∠ABD = ∠ABC + ∠CBD
∠ACD = ∠ACB + ∠BCD
But ∠ABC = ∠ACB and ∠CBD = ∠BCD
Therefore,
∠ABD = ∠ACD
Q6. 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.
Given: AB = AC and AD = AB
So,
AB = AC = AD
Hence, A is equidistant from B, C and D.
Therefore, points B, C and D lie on a circle with centre A.
Now B, A, D are collinear, so BD is a diameter of this circle.
Angle subtended by a diameter at the circumference is 90°.
Therefore,
∠BCD = 90°
Q7. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Given: ∠A = 90° and AB = AC
Since AB = AC, triangle ABC is isosceles.
Therefore,
∠B = ∠C
Now,
∠A + ∠B + ∠C = 180°
90° + ∠B + ∠C = 180°
∠B + ∠C = 90°
But ∠B = ∠C
So,
2∠B = 90°
∠B = 45°
∠C = 45°
Answer: ∠B = 45°, ∠C = 45°
Q8. Show that the angles of an equilateral triangle are 60° each.
Given: Let ΔABC be an equilateral triangle.
So,
AB = BC = CA
Since AB = AC,
∠B = ∠C
Since AB = BC,
∠A = ∠C
Thus,
∠A = ∠B = ∠C
Now,
∠A + ∠B + ∠C = 180°
So,
3∠A = 180°
∠A = 60°
Similarly,
∠B = 60° and ∠C = 60°
Hence, each angle of an equilateral triangle is 60°.