## Triangles

**Chapter 7**

**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,

AD = DA [Common]

∠ADB = ∠ADC [Each 90°]

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)

AB = AD [Given] …(2)

From (1) and (2), we have

AC = AD

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,

∠ADC + ∠ACD + ∠CAD = 180°

⇒ 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.