Exercise: 7.2 (Page No: 123)
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
Solution:
Given:
AB = AC and
the bisectors of B and C intersect each other at O
(i) Since ABC is an isosceles with AB = AC,
B = C
½ B = ½ C
⇒ OBC = OCB (Angle bisectors)
∴ OB = OC (Side opposite to the equal angles are equal.)
(ii) In ΔAOB and ΔAOC,
AB = AC (Given in the question)
AO = AO (Common arm)
OB = OC (As Proved Already)
So, ΔAOB ΔAOC by SSS congruence condition.
BAO = CAO (by CPCT)
Thus, AO bisects A.
2. In ΔABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that ΔABC is an isosceles triangle in which AB = AC.
Solution:
It is given that AD is the perpendicular bisector of BC
To prove:
AB = AC
Proof:
In ΔADB and ΔADC,
AD = AD (It is the Common arm)
ADB = ADC
BD = CD (Since AD is the perpendicular bisector)
So, ΔADB ΔADC by SAS congruency criterion.
Thus,
AB = AC (by CPCT)
3. 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:
(i) BE and CF are altitudes.
(ii) AC = AB
To prove:
BE = CF
Proof:
Triangles ΔAEB and ΔAFC are similar by AAS congruency since
A = A (It is the common arm)
AEB = AFC (They are right angles)
AB = AC (Given in the question)
∴ ΔAEB ΔAFC and so, BE = CF (by CPCT).
4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that
(i) ΔABE ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.
Solution:
It is given that BE = CF
(i) In ΔABE and ΔACF,
A = A (It is the common angle)
AEB = AFC (They are right angles)
BE = CF (Given in the question)
∴ ΔABE ΔACF by AAS congruency condition.
(ii) AB = AC by CPCT and so, ABC is an isosceles triangle.
5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ABD = ACD.
Solution:
In the question, it is given that ABC and DBC are two isosceles triangles.
We will have to show that ABD = ACD
Proof:
Triangles ΔABD and ΔACD are similar by SSS congruency since
AD = AD (It is the common arm)
AB = AC (Since ABC is an isosceles triangle)
BD = CD (Since BCD is an isosceles triangle)
So, ΔABD ΔACD.
∴ ABD = ACD by CPCT.
6. 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:
It is given that AB = AC and AD = AB
We will have to now prove BCD is a right angle.
Proof:
Consider ΔABC,
AB = AC (It is given in the question)
Also, ACB = ABC (They are angles opposite to the equal sides and so, they are equal)
Now, consider ΔACD,
AD = AB
Also, ADC = ACD (They are angles opposite to the equal sides and so, they are equal)
Now,
In ΔABC,
CAB + ACB + ABC = 180°
So, CAB + 2ACB = 180°
⇒ CAB = 180° – 2ACB — (i)
Similarly, in ΔADC,
CAD = 180° – 2ACD — (ii)
also,
CAB + CAD = 180° (BD is a straight line.)
Adding (i) and (ii) we get,
CAB + CAD = 180° – 2ACB+180° – 2ACD
⇒ 180° = 360° – 2ACB-2ACD
⇒ 2(ACB+ACD) = 180°
⇒ BCD = 90°
7. ABC is a right-angled triangle in which A = 90° and AB = AC. Find B and C.
Solution:
In the question, it is given that
A = 90° and AB = AC
AB = AC
⇒ B = C (They are angles opposite to the equal sides and so, they are equal)
Now,
A+B+C = 180° (Since the sum of the interior angles of the triangle)
∴ 90° + 2B = 180°
⇒ 2B = 90°
⇒ B = 45°
So, B = C = 45°
8. Show that the angles of an equilateral triangle are 60° each.
Solution:
Let ABC be an equilateral triangle as shown below:
Here, BC = AC = AB (Since the length of all sides is same)
⇒ A = B =C (Sides opposite to the equal angles are equal.)
Also, we know that
A+B+C = 180°
⇒ 3A = 180°
⇒ A = 60°
∴ A = B = C = 60°
So, the angles of an equilateral triangle are always 60° each.