# GSEB Solutions Class 9 Maths Chapter 7 Triangles Ex 7.2

Gujarat Board GSEB Solutions Class 9 Maths Chapter 7 Triangles Ex 7.2 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 7 Triangles Ex 7.2

Question 1.
In an isosceles triangle ABC, with AB = AC, the bisectors of âˆ B and âˆ C intersect each other at O. Join A to O. Show that.
1. OB = OC
2. AO bisects âˆ A
Solution:
1. AB = AC (given)
âˆ B = âˆ C
(Angle opposite to equal sides are equal)

âˆ B = âˆ C (Dividing by 2)
(OB and OC angle bisectors of âˆ B and âˆ C)
â‡’ âˆ 1 = âˆ 2
Hence OB = OC
(Sides opposite to equal angles are equal)

2. Now in Î”AOB and Î”AOC
AB = AC (given)
âˆ B = âˆ C (Angle opposite to equal sides are equal)
â‡’$$\frac {1}{2}$$âˆ B = $$\frac {1}{2}$$âˆ C (Dividing by 2)
â‡’âˆ 3 = âˆ 4
and OB = OC (Proved above)
Hence âˆ AOB â‰… âˆ AOC (by SAS congruency)
âˆ´ âˆ OAB = âˆ OAC (By CPCT)
Hence AO bisects âˆ A.

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

Solution:
Given: Î”ABC in which AD âŠ¥ BC
and BD = CD
To Prove: Î”ABC is an isosceles triangle in which
AB = AC
BD = CD (AD is lar bisector of BC)
âˆ ADB = âˆ ADC (each 90Â°)
âˆ´ Î”ADB â‰… Î”ADC (By SAS congruency)
âˆ´ AB = AC (by CPCT)
Therefore ABC is an isosceles triangle in which AB = AC.

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

Solution:
Given: ABC is an isosceles triangle in which BE and CF are altitudes on AC and AB respectively.
To Prove:
BE = CF
Proof: In Î”ABE and Î”ACF,
AB = AC (Equal sides of an isosceles triangle)
âˆ AEB = âˆ AFC (each 90Â°)
âˆ BAE = âˆ CAF (Common)
âˆ´ Î”AEB â‰… Î”ACF (By AAS congruency)
Hence BE = CF (by CPCT)

Question 4.
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) Î”ABE â‰… Î”ACF
(ii) AB = AC (i.e. ABC is an isosceles triangle

Solution:
Given: Î”ABC in which BE = CF.
(i) In Î”ABE and Î”ACF,
âˆ BAE = âˆ CAF (Common)
âˆ AEB = âˆ AFC (Each 900)
BE = CF (given)
âˆ´ Î”ABE â‰… Î”ACF (By AAS congruency)

(ii) Î”ABE â‰… Î”ACF (Proved above in part i)
âˆ´ AB = AC (by CPCT)
Î”ABC is an isosceles triangle.

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

Solution:
ABC is an isosceles triangle.
AB = AC (given)
âˆ ABC = âˆ ACB ……..(1) (Angles opposite to equal sides are equal)
Now DBC is an isosceles triangle.
BD = CD (given)
âˆ´ âˆ DBC = âˆ DCB (Angles opposite to equal sides are equal)
Adding equation (1) and (2), we get
âˆ ABC + âˆ DBC = âˆ ACB + âˆ DCB
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 figure). Show that âˆ BCD is a right angle.
Or
In the given figure AB = AC and AB = AD, Prove that âˆ BCD = 90Â°.

Solution:
Î”ABC is an isosceles triangle.
AB = AC (given)
âˆ´ âˆ ABC = âˆ ACB
(Angles opposite to equal sides are equal)
In Î”ACD,
AB = AD (by construction)
But AB = AC
âˆ´ AD = AC
âˆ ADC = âˆ ACD …….(2)
(Angles opposite to equal sides are equal)
Adding eqn. (1) and (2), we get
âˆ ABC + âˆ ADC = âˆ ACB + âˆ ACD
âˆ ABC + âˆ ADC = âˆ BCD
âˆ ABC + âˆ BDC = âˆ BCD ………(3)
(âˆ ADC = âˆ BDC)
Now in âˆ ABC,
âˆ ABC + âˆ BDC + âˆ BCD = 180Â° (By Angle Sum Property)
âˆ BCD + âˆ BCD = 180Â°
[From eqn. (3)]
2 âˆ BCD = 180Â°
âˆ BCD = $$\frac {180Â°}{2}$$
âˆ BCD = 90Â°

Question 7.
ABC is a right-angled triangle in which âˆ A = 90Â°, and AB = AC. Find âˆ B and âˆ C.
Solution:
In Î”ABC,
AB = AC
âˆ B = âˆ C
(Angles opposite to equal sides are equal)

Now in âˆ ABC,
âˆ A + âˆ B + âˆ C = 180Â° (by angle sum property)
90Â°+ âˆ C + âˆ C = 180Â° (âˆ B = âˆ C)
2âˆ C = 180Â° – 90Â°
2âˆ C = 90Â°
âˆ C = 45Â°
âˆ B = âˆ C = 45Â°

Question 8.
Show that the angles of an equilateral triangle are 60Â° each.
Solution:
Given: ABC is an equilateral triangle.
To Show:
âˆ A = âˆ B = âˆ C = 60Â°

Proof: ABC is an equilateral triangle.
âˆ´ AB = AC
âˆ´ âˆ C = âˆ B …….(1) (Angles opposite to equal sides are equal)
BC = AC
âˆ A = âˆ B ………(2) (Angles opposite to equal sides are equal)
From eqn. (1) and (2), we get
âˆ A = âˆ B = âˆ C = x (Say)
Now in Î”ABC,
âˆ A + âˆ B + âˆ C = 180Â°
x + x + x = 180Â°
3x = 180Â°
x = 60Â°
Hence âˆ A + âˆ B = âˆ C = 60Â°
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