Triangles – Exercise 7.2 – (MATHEMATICS) – 9th Class

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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

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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.

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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.


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Sol. – Since AD is perpendicular bisector of BC.
∴   BD = CD and ΔADB = ΔADC = 90°                                      …(i)

Consider triangles ADB and ADC,

We have BD = DC

and. ∠ADB = ∠ADC                                                              [From (i)]

AD = AD                                                                                 [Common]

∴    ΔADB ≅ ΔADC                                                               [SAS rule]

∴      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.


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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

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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.

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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.


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

BD = CD                                            [ Given ]

AD = AD                                            [ Common ]

∴     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.


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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.

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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.

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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°.


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