# Triangles : Exercise 7.2 (Mathematics NCERT Class 9th)

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**Q.1Â Â Â Â In an isosceles triangle ABC, with AB = AC, the bisectors of intersect each other at O. Join A to O. Show that : **

Â Â Â Â Â Â Â Â Â Â Â Â Â **(i) OB = OC Â Â Â Â Â Â Â Â Â (ii) AO bisects .**

**Sol.**

(i) In ABC, we have

AB = AC

[Since angles opposite to equal sides are equal]

[Since OB and OC bisect B and C respectively. Therefore ]

OB = OC [Since sides opp. to equal are equal]

(ii) Now, in ABO and ACO , we have

AB = AC [Given]

[From (1)]

OB = OC [From (2)]

Therefore By SAS criterion of congruence, we have

[Since corresponding parts of congruent triangles are equal]

AO bisects .

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

In ABD and ACD, we have

DB = DC [Given]

[since AD BC]

AD = AD [Common]

Therefore by SAS criterion of congruence, we have.

AB = AC

[Since corresponding parts of congruent triangles are equal]

Hence, ABC is isosceles.

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

In ABE and ACF, we have

[Since Each = 90Âº]

[Common]

and BE = CF [Given]

Therefore By AAS criterion of congruence, we have

AB = AC

[Since Corresponding parts of congruent triangles are equal]

Hence, ABC is isosceles.

**Q.4Â Â Â Â Â ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that Â Â Â Â Â Â Â Â Â Â Â **

**(i) Â Â Â Â Â Â Â Â Â Â Â**

Â Â Â Â Â Â Â Â Â Â Â

Â Â Â Â Â Â Â Â Â Â Â

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

Let BE AC and CF AB.

In ABE and ACF, we have

[Since Each = 90Âº]

[Common]

and , AB = AC [Given]

Therefore By AAS criterion of congruence,

BE = CF

[Since corresponding parts of congruent triangles are equal]

**Q.5Â Â Â ABC and DBC are two isosceles triangles on the same base BC (see figure). Show that . **

In ABC, we have

AB = AC

... (1)

[Since angles opposite to equal sides are equal]

In , we have

BD = CD

... (2)

[Since angles opposite to equal sides are equal]

Adding (1) and (2), we have

**Q.6Â Â Â Â Â ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see figure). Show that is a Â Â ****right angle. **

In ABC, we have

AB = ACÂ Â Â [given]

... (1)

[Since angles opp. to equal sides are equal]

Now, AB = AD [Given]

Therefore AD = AC [Since AB = AC]

Thus , in ADC, we have

AD = AC

... (2)

[Since angles opp. to equal sides are equal]

Adding (1) and (2) , we get

[Since ]

Â [Adding on both sides]

Â [Angle sum proprty]

90Âº

Hence , is a right angle.

**Q.7Â Â Â Â ABC is a right angled triangle in which 90Âº and AB = AC. Find .**

**Sol. **

We have,

90Âº

AB = AC

[Since angles opp. to equal sides of a triangle are equal]

Also 180Âº [Angle - sum property]

90Âº + 180Âº [Since ]

180Âº â€“ 90Âº = 90Âº

Therefore

**Q.8Â Â Â Â Â Show that the angles of an equilateral triangle are 60Âº each.**

**Sol. **

Let ABC be an equilateral triangle so that AB = AC = BC.

Now Since AB = AC

... (1)Â [Since angles opp. to equal sides are equal]

Also since, CB = CA

.... (2)Â [Since angles opp. to equal sides are equal]

From (1) and (2), we have

Also 180Âº [Angle - sum property]

Thererfore 180Âº

180Âº

60Âº

Therefore 60Âº

Thus , each angle of an equilateral triangle is 60Âº.