0

Exercise 7.2

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

Sol. –
In  ΔABC, AB = AC                                              [Given]

⇒   ∠ABC = ∠ACB                                               [Angles opposite to equal sides are equal.]

$\Rightarrow {1 \over 2}\angle ABC = {1 \over 2}\angle ACB$

⇒    ∠OBC = ∠OCB                                    [ OB and OC are bisectors of LABC and ∠ACB respectively.]

⇒    OC = OB                                                       …(i)

[Sides opposite to equal angles of a triangle are equal.]

(ii) AO bisects ∠A.

Sol. –
Consider triangles  AOB and AOC,

We have      OC = OB                                                      [From (i)]

AO = OA                                                                           [Common]

and AB = AC                                                                    [Given]

∴  ΔAOB  ≅  ΔAOC                                                         [SSS rule]

⇒       ∠OAB = ∠OAC                                                   [CPCT]

⇒        AO bisects ∠ BAC.

2. In ΔABC, AD is the perpendicular bisector of BC (see figure). Show that ΔABC is an isosceles triangle in which AB = AC.

Sol. – Since AD is perpendicular bisector of BC.

We have BD = DC

∴      AB = AC.
Therefore, ΔABC is an isosceles triangle with AB = AC.           [CPCT]

3.  ABC is an isosceles triangle in which  altitudes BE and CF are drawn to equal sides AC and AB respectively (see figure). Show that these altitudes are equal.

Sol.  In ΔABC,   AB = AC                           [Given]

⇒      ∠ACB = ∠ABC                                      ….(i)

[Angles opposite to equal sides of a Consider triangles are equal. ]

Consider triangle BFC and BCE,

We have    ∠FBC = ∠ECB                                          [From (i)]

BC = CB                                                                        [Common]

∠BFC = ∠CEB                                                             [90° each]

∴      ΔBCF ≅  ΔCBE                                                   [AAS rule]

⇒                CF = BE.

Hence, altitudes to the equal sides of a triangle are equal.

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that:
(i) ΔABE ≅ ΔACF

Sol. –

Consider triangles ABE and ACF,
We have BE = CF                                                             [Given]

∠A = ∠A                                                                             [Common]

∠AEB = ∠AFC                                                                  [90° each]

∠ ABE ≅  ∠ACF                                                               [AAS rule]

(ii) AB = AC, i.e., ABC is an isosceles triangle.

Sol. –

Since ΔABE ≅ ΔACF                                    [Proved above]

Hence,     AB = AC

i.e., Δ ABC is isosceles.

5. ΔABC and DBC are two isosceles triangles on same base BC (see figure). Show that ΔABD = ΔACD.

Sol.
Construction: Join A and D.
Proof : Consider triangles ABD and ACD,
We have AB = AC                            [Given ]

BD = CD                                            [ Given ]

∴     ΔABD ≅ ΔACD                         [ SSS rule ]

∴     ∠ABD = ∠ACD.                         [ CPCT ]

6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that ∠BCD is a right angle.

Sol.  As   AB = AC                                          [Given]

⇒            ∠1 = ∠2                                                   …(i)

[Angles opposite to equal sides of a triangle are equal.]

and    AC = AD                            ($AB = AD$)         [ Given ]

⇒    ∠3 = ∠4                                                                    …(ii)

[Angles opposite to equal sides of a triangle are equal.]

Also, in ΔDBC,
∠DBC + ∠BCD + ∠CDB = 180°

[Sum of angles of a triangle is 180°.]

⇒ ∠1 + (∠2 + ∠3) + ∠4 = 180°

⇒  ∠2 + ∠2 + ∠3 + ∠3   =180°                   [Using (i), (ii)]

⇒    2(∠2 + ∠3 )    =   180°

⇒        ∠ 2      +    ∠ 3   =    90°     ⇒     ∠BCD = 90°

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

Sol.

As   AB = AC                                             [Given]

⇒ ∠B = ∠C               …(i)                             [Angles opposite to equal sides of a triangle are equal.]

In  ΔABC , ∠A + ∠B + ∠C  =  180°           [Sum of angles of a triangle is 180°.]

⇒   90°+   ∠B   +   ∠B  =   180°

⇒       2∠B   =    90°                                      [From (i)]

⇒           B = 45°

∴             ∠B = ∠C= 45°.

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

Sol.   As ΔABC is equilateral.

∴    AB = BC = CA
Now.               AB = BC   ⇒  ∠C = ∠A     ….(i)

[Angles opposite to equal sides of 6. triangle are equal ]

Similarly, BC = AC  ⇒ ∠A = ∠B             [Reason same as above ]

⇒       ∠A  =   ∠B   = ∠C               ..(iii) [From (i) and (ii) ]

In Δ ABC , ∠A + ∠B +∠C    = 180°                             [Sum angles of  a triangle is 180°]

⇒ ∠A + ∠A  + ∠A  = 180°                                          [ From (iii)]

⇒         3∠A = 180°

=            ∠A =  60°

∴   ∠A = ∠B = ∠C= 60°.