# Triangles (Exercise 7.2)

## EX 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:

(i) OB = OC

(ii) AO bisects A

Solution:
i) In ∆ABC,
AB = AC                   [Given]
∴ ∠ABC = ∠ACB     [Angles opposite to equal sides are equal]

⇒ ½∠ABC = ½∠ACB
or ∠OBC = ∠OCB
⇒ OC = OB           [Sides opposite to equal angles of a ∆ are equal]

(ii) In ∆ABO and ∆ACO,
AB = AC                  [Given]
∠OBA = ∠OCA       [ ∵∠B = ∠C]
OB = OC                 [Proved above]
∆ABO ≅ ∆ACO       [By SAS congruency]
⇒ ∠OAB = ∠OAC    [By C.P.C.T.]
⇒ AO bisects ∠A.

## EX 7.2 QUESTION 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:

BD = CD                   [AD is the bisector of BC.]
Now, in ∆ABD and ∆ACD,
BD = CD                  [Proved above]
∴ ∆ABD ≅ ∆ACD     [By SAS congruency]
⇒ AB = AC               [By C.P.C.T.]
Thus, ∆ABC is an isosceles triangle.

## EX 7.2 QUESTION 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:

In ∆ABE and ∆ACF
∠ABE = ∠ACF        [Each 90°]
∠A = ∠A                [Common]
AB = AC                [Given]
∆ABE ≅ ∆ACF        [By AAS congruency]
∠BCE = ∠CBF [Proved above]
BE = CF                  [By C.P.C.T.]

## EX 7.2 QUESTION 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:

(i) In ∆ABE and ∆ACE,
∠AEB = ∠AFC            [Each 90° ]
∠A = ∠A                    [Common]
BE = CF                     [Given]
∴ ∆ABE ≅ ∆ACF        [By AAS congruency]

(ii) Since, ∆ABE ≅ ∆ACF
∴ AB = AC                [By C.P.C.T.]
⇒ ABC is an isosceles triangle.

## EX 7.2 QUESTION 5.

ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ABD = ACD.

Solution:

In ∆ABC,
AB = AC            [ABC is an isosceles triangle]
∴ ∠ABC = ∠ACB …(1)         [Angles opposite to equal sides of a ∆ are equal]
In ∆DBC,
DB = DC                               [BDC is an isosceles triangle]
∴ ∠DBC = ∠DCB …(2)          [Angles opposite to equal sides are equal]
Adding (1) and (2), we have
∠ABC + ∠DBC = ∠ACB + ∠DCB
⇒ ∠ABD = ∠ACD.

## EX 7.2 QUESTION 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:

AB = AC [Given] …(1)
From (1) and (2), we have
Now, in ∆ABC, we have
∠ABC + ∠ACB + ∠BAC = 180° [Angle sum property of a A]
⇒ 2∠ACB + ∠BAC = 180° …(3)
[∠ABC = ∠ACB (Angles opposite to equal sides of a A are equal)]
Similarly, in ∆ACD,
⇒ 2∠ACD + ∠CAD = 180° …(4)
[∠ADC = ∠ACD (Angles opposite to equal sides of a A are equal)]
Adding (3) and (4), we have
2∠ACB + ∠BAC + 2 ∠ACD + ∠CAD = 180° +180°
⇒ 2[∠ACB + ∠ACD] + [∠BAC + ∠CAD] = 360°
⇒ 2∠BCD +180° = 360° [∠BAC and ∠CAD form a linear pair]
⇒ 2∠BCD = 360° – 180° = 180°
⇒ ∠BCD = 180°/2 = 90°
Thus, ∠BCD = 90°

## EX 7.2 QUESTION 7.

ABC is a right-angled triangle in which A = 90° and AB = AC. Find B and C.

Solution:

In ∆ABC, we have AB = AC                  [Given]
∴ Their opposite angles are equal.

⇒ ∠ACB = ∠ABC
Now, ∠A + ∠B + ∠C = 180° [Angle sum property of a ∆]
⇒ 90° + ∠B + ∠C = 180°                [∠A = 90°(Given)]
⇒ ∠B + ∠C= 180°- 90° = 90°
But ∠B = ∠C
∠B = ∠C = 90°/2 = 45°
Thus, ∠B = 45° and ∠C = 45°

## EX 7.2 QUESTION 8.

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

Solution:

In ∆ABC, we have

AB = BC = CA                                    [ABC is an equilateral triangle]
AB = BC
⇒ ∠A = ∠C …(1)                                 [Angles opposite to equal sides of a A are equal]
Similarly, AC = BC
⇒ ∠A = ∠B …(2)
From (1) and (2), we have
∠A = ∠B = ∠C = x (say)
Since, ∠A + ∠B + ∠C = 180°            [Angle sum property of a A]
∴ x + x + x = 180o
⇒ 3x = 180°
⇒ x = 60°
∴ ∠A = ∠B = ∠C = 60°
Thus, the angles of an equilateral triangle are 60° each.