NCERT Solutions For Class 9 Math Chapter – 7 Exercise – 7.2
Q1. 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:
- OB = OC (ii) AO bisects ∟A.
Solution:-
In ∆ABC , AB = AC
∟ABC = ∟ACB [ Angles opposite to equal sides are equal].
= 1/2 ∟ABC = 1/2∟ACB
∟OBC = ∟OCB [ OB and OC are bisectors of ∟ABC and ∟ACB respectively.]
OC = OB [Sides opposite to equal angles of a triangle are equal.]
- Consider triangles AOB and AOC,
OC = OB [From (I)]
AO = OA [Common]
∆AOB ∆AOC [SSS rule]
∟OAB = ∟OAC [CPCT]
= AO bisects ∟BAC.
Q2. In ∆ABC , AD is the perpendicular bisector of BC Show that ∆ABC is an isosceles triangle in which AB = AC.
Solution:-
In triangles ADB and ADC,
BD = BC
∟ADB = ∟ADC
AD = AD
∆ADB ∆ADC [SAS Rule]
AB = AC.
Therefore ∆ABC is an isosceles triangle with AB = AC. [CPCT]
Q3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively Show that these altitudes are equal.
Solution:-
In ∆BFC and BCE.
∟FBC = ∟ECB.
BC = CB
∟BFC = ∟CEB
∆BCF ∆CBE.
CF = BE.
Q4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal Show that:
- ∆ABE ∆ACF
- AB = AC ABC is an isosceles triangle.
Solution:-
- In triangle ABE and ACF,
BE = CF
∟A = ∟A
∟AEB = ∟AFC
∆ABE ∆ACF [AAS Rule]
- Since ∆ABE ∆ACF
Hence AB = AC
∆ABC is a isosceles.
Q5. ABC and DBC are two isosceles triangles on same base BC Show that ∟ABD = ∟ACD
Solution:-
In triangle ABD and ACD,
AB = AC
BD = CD
AD = AD
∆ABD ∆ACD
∟ABD = ∟ACD
Q6. ∆ABC is an isosceles triangle in which AB = AC Side BA is produced to D Such that AD = AB Show that ∟BCD is a right angle.
Solution:-
AB = AC [Given]
∟1 = ∟2 (I)
[Angles opposite to equal sides of a triangle are equal.]
AB = AD (Given)
And
AC = AD
∟3 = ∟4
[Angles opposite to equal sides of a triangle are equal.]
In ∆ DBC,
∟DBC + ∟BCD + ∟CDB = 1800
[ Sum of angles of a triangle is 1800]
= ∟1 + ∟2 + ∟3 + ∟4 = 1800
= ∟2 + ∟2 + ∟3 + ∟3 = 1800
2(∟2 + ∟3) = 1800
∟2 + ∟3 = 900
Q7. ABC is a right angled triangle in which ∟A = 900 and AB = AC Find ∟B and ∟C.
Solution:-
As AB = AC
∟B = ∟C [ Angels opposite to equal sides of a triangle are equal].
In ∆ABC ,
∟A + ∟B + ∟C = 1800
2∟B = 900
∟B = 450
∟B = ∟C = 450
Q8. Show that the angels of an equilateral triangle are 600 each.
Solution:-
In ∆ABC is equilateral
AB = BC = CA
AB = BC
∟A = ∟C
[ Angles opposite to equal sides of a triangle are equal].
BC = AC = ∟A = ∟B
∟A = ∟B = ∟C
In ∆ABC ,
∟A + ∟B + ∟C = 1800
[ Sum of angels of a triangle is 1800.]
= ∟A + ∟A + ∟A = 1800
3∟A = 1800
∟A = 600
∟A = ∟B = ∟C = 600