# Class 9 NCERT Solutions- Chapter 7 Triangles – Exercise 7.2

### (i) OB = OC                     (ii) AO bisects âˆ A

Solution:

Given: (i) An isosceles âˆ†ABC in which AB=AC

(ii) bisects of âˆ B and âˆ C each other at O.

Show: (i) OB=OC

(ii) AO bisects âˆ A (âˆ 1=âˆ 2)

(i) In âˆ†ABC,

AB = AC

âˆ B =âˆ C                          [angles opposite to equal sides are equal]

1/2 âˆ B = 1/2âˆ C

âˆ OBC=âˆ OCB

âˆ´OB = OC                      [sides opposite equal âˆ  are equal]

AB = AC                         [given side]
1/2 âˆ B = 1/2âˆ C
âˆ ABO = âˆ ACO               [Angle]
BO = OC                        [proved above side]
âˆ´âˆ†AOB â‰… AOC
Thus âˆ 1 = âˆ 2
Therefore, AO bisects âˆ A

### Question 2. In Î”ABC, AD is the perpendicular bisector of BC (see fig.). Show that Î”ABC is an isosceles triangle in which AB = AC.

Solution:

Given: AD is âŠ¥ bisector of BC

Show: AB=BC

In âˆ†ABD and âˆ†ACD

BD=DC                    [AD is âŠ¥ bisector side]

âˆ´âˆ†ABDâ‰…âˆ†ACD        [S.A.S]

AB=AC                     [C.P.C.T]

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

Solution:

Given: AB=AC,BE and CF are altitudes

Show: BE=CF

In âˆ†AEB and âˆ†AFC,

âˆ E=âˆ F                     [Each 90Â° angle]

âˆ A=âˆ A                    [common angle]

AB=AC                    [given side]

âˆ´âˆ†AEBâ‰…âˆ†AFC        [A.A.S]

BE=CF                    [C.P.C.T]

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

Solution:

Given: Altitudes BE and CF to sides AC and AB are equal

Show: (i) Î”ABE â‰… Î”ACF

(ii) AB = AC

(i) In âˆ†ABF and âˆ†ACF,

âˆ E=âˆ F                    [Each 90Â° angle]

âˆ A=âˆ A                   [common angle]

AB=AC                   [given] S

âˆ´âˆ†AEBâ‰…âˆ†AFC       [A.A.S]

(ii) AB=AC             [C.P.C.T]

### Question 5. ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that âˆ ABD = âˆ ACD.

Solution:

Given: AB=AC,BD=DC

Show: âˆ ABD = âˆ ACD

In âˆ†ABD,

âˆ´âˆ 1=âˆ 2                  [angle opposite to equal sides are equal] [1]

In âˆ†BDC,

BD=DC

âˆ´âˆ 3=âˆ 4                  [angle opposite to equal sides are equal] [2]

âˆ 1+âˆ 2= âˆ 2+âˆ 4

âˆ ABD=âˆ ACD

### Question 6. Î”ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig.). Show that âˆ BCD is a right angle.

Solution:

In âˆ†ABC,

AB=AC

âˆ ACB=âˆ ABC           [1]

In âˆ†ACD,

âˆ BCD=âˆ ABC+âˆ BDC

Adding âˆ BCD on both side

âˆ BCD+âˆ BCD=âˆ ABC+âˆ BDC+âˆ BCD

2âˆ BCD=180Â°

âˆ BCD=(180Â°)/2=90Â°

### Question 7. ABC is a right-angled triangle in which âˆ A = 90Â° and AB = AC. Find âˆ B and âˆ C.

Solution:

Find: âˆ B=? and âˆ C?

In âˆ†ABC,

AB=AC

âˆ´âˆ B=âˆ C                           [angle opposite to equal side are equal]

âˆ A+âˆ B+âˆ C=180Â°            [angle sum property of triangle]

90Â°+âˆ B+âˆ B=180Â°

2âˆ B=180Â°-90Â°

âˆ B=(90Â°)/2=45Â°

Therefore, âˆ B=45Â° and âˆ C =45Â°

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

Solution:

Given: Let âˆ†ABC is an equilateral âˆ†

Show: âˆ A=âˆ B=âˆ C=60Â°

In âˆ†ABC,

AB=AC

âˆ B=âˆ C                    [1]

Also

AC=BC

âˆ B=âˆ A                    [2]

From 1,2 and 2

âˆ A=âˆ B=âˆ C

In âˆ†ABC,

âˆ A+âˆ B+âˆ C=180Â°      [angle sum property of triangle]

âˆ A+âˆ A+âˆ A=180Â°

3âˆ A=180Â°

âˆ A=(180Â°)/3=60Â°

âˆ A=60Â°

âˆ´âˆ B=60Â° and âˆ C=60Â°

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