Triangles-9th-Math: 9 Math
Solution of NCERT Exercise 7.2: 9th math
Question (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
(ii) OA bisects ∠ A
Solution
Given, Δ ABC where AB = AC
OB bisects ∠ B
OC bisects ∠ C
To show
(i) OB = OC
(ii) OA bisects ∠ A
Proof (i)
In Δ ABC
OC bisects ∠ C
Therefore, ∠ ACO = ∠ OCB
And OB bisects ∠ B
Therefore, ∠ ABO = ∠ OBC
Since, in an isosceles triangle opposite angles of equal sides are equal
Therefore, ∠ ABC = ∠ ACB
⇒ ∠ ACO = ∠ ABO ---------- (i)
Now, in Δ AOC and Δ AOB
AO = AO (common side in both of the triangle)
∠ ACO = ∠ ABO (From equation (i))
AB = AC (As given in question)
Therefore, by SAS congruence criterion
Δ AOC ≅ Δ AOB
Therefore, by CPCT (Corresponding parts of congruence triangles)
OC = OB Proved
Proof (ii)
Since, Δ AOC ≅ Δ AOB
Therefore, by CPCT (Corresponding parts of congruence triangles)
∠ CAO = ∠ BAO
⇒ AO bisects angle A. Proved
Question (2) In Δ ABC, AD is the perpendicular bisector of BC (see figure). Show that Δ ABC is an isosceles triangle in which AB = AC.
Solution
Given, In Δ ABC
AD is perpendicular bisector of BC.
To show
Δ ABC is an isosceles triangle and AB = AC
Proof
In Δ ABD and Δ ADC
BD = DC (Because as given in question AD is bisector of BC)
∠ ADB = ∠ ADC = 900
(AD is perpendicular bisector of BC as given in question)
AD = AD (common side in both of the triangle)
Therefore, by SAS (Side-Angle-Side) congruence criterion
Δ ABD ≅ Δ ADC
Therefore, by CPCT (Corresponding parts of congruence triangles)
AB = AC
Hence, Δ ABC is an isosceles triangle in which AB = AC Proved
Question (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.
Solution
Given, Δ ABC is a isosceles triangle
And BE and CF are drawn to equal sides AC and AB
i.e. AC = AB (Equal sides of given triangle)
To prove CB = BE
Proof
In Δ BFC and Δ BEC
∠ FBC = ∠ ECB
(Δ ABC is an isosceles triangle therefore, opposite angles of equal sides are equal)
BC = BC (Common sides in both of the triagnles)
∠ BFC = ∠ BEC = 900
(CF and BE are perpendiculars on AC and AB respectively)
Therefore, by ASA (Angle-Side-Angle) congruence criterion
Δ BFC ≅ Δ BEC
Therefore, by CPCT (Corresponding parts of congruence triangles)
CF = BE
i.e. altitudes of the given triangles are equal. Proved
Question (4) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see figure). Show that
(i) Δ ABE ≅ Δ ACF
(ii) AB = AC, i.e. ABC is an isosceles triangle
Same figure as in question (3)
Solution
Given, In Δ ABC
CF and BE are altitude on AB and AC respectively
AB = AC (Equal sides of isosceles triangle)
To show that,
(i) Δ ABE ≅ Δ ACF
(ii) AB = AC, i.e. ABC is an isosceles triangle
Proof (i)
In Δ ABE and Δ ACF
∠ BAE = ∠ CAF (common side of the triangles)
AB = AC (As given in question)
∠ CFA = ∠ BEA = 900
(Because CF and BE are altitude on AB and AC respectively, as given in question)
Thus, by ASA (Angle-Side-Angle) congruence criterion
Δ ABF ≅ Δ ACF Proved
Proof (ii)
As given in question, AC and AB are equal
Thus, AB = AC, i.e. Δ ABC is an isosceles triangle Proved
Question (5) ABC and EBC are two isosceles triangles on the same base BC (see figure). Show that ∠ ABD = ∠ ACD.
Solution
Given, Δ ABC and Δ DCB are isosceles triangle
AB = AC and BD = DC
Both of the triangles have same base BC
To show ∠ ABD = ∠ ACO
Construction
A and D are joined.
Proof
In Δ ABD and Δ ACD
AB = AC (As given in question)
BD = DC (As given in question)
AD = AD (Common side in both of the triangles)
Therefore, by SSS (Side-Side-Side) congruence criterion
Δ ABD ≅ Δ ADC
Therefore, by CPCT (Corresponding parts of congruence triangles)
∠ ABD = ∠ ACD Proved
Question (6) Δ ABC 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.
Solution
Given, Δ ABC is an isosceles triangle
AB = AC = AD
To show ∠ BCD is right angle, i.e. ∠ BCD = 900
Proof
In Δ ABC
AB = AC
⇒ ∠ ABC = ∠ ACB ----------- (i)
(Angles opposite to the equal sides of a triangle are equal)
Similarly, in Δ ACD
AC = AD
Therefore, ∠ ACD = ∠ ADC --------- (ii)
(Angles opposite to the equal sides of a triangle are equal)
Now, in Δ ABC
∠ CAB + ∠ ACB + ∠ ABC = 1800
(Because sum of all angles of a triangle is equal to 1800)
⇒ ∠ CAB + ∠ ACB + ∠ ACB = 1800
[Because from equation (i) ∠ ABC = ∠ ACB]
⇒ ∠ CAB + 2 ∠ ACB = 1800
⇒ ∠ CAB = 1800 – 2 ∠ ACB ------------- (iii)
Simiarly, in Δ ADC
∠ CAD = 1800 – 2 ∠ ACD ------------- (iv)
Now, by adding equation (iii) and (iv), we get
∠ CAB + ∠ CAD = 1800 – 2 ∠ ACB + 1800 – 2 ∠ ACD
⇒ 1800 = 1800 + 1800 – 2 ∠ ACB – 2 ∠ ACD
[Because, ∠ CAB + ∠ CAD = 1800]
⇒ 1800 = 3600 – 2(∠ ACB + ∠ ACD)
⇒ 2 (∠ ACB + ∠ ACD) = 3600 – 1800
⇒ 2( ∠ ACB + ∠ ACD) = 1800
[Because, ∠ ACB + ∠ ACD = BCD]
⇒ 2 ∠ BCD = 1800
⇒ ∠ BCD = 1800 / 2
⇒ ∠ BCD = 900 Proved
Question (7) ABC is a right angled triangle in which ∠ A = 900 and AB = AC. Find ∠ B and ∠ C.
Solution
Given, ABC is a right angle triangle
In which, ∠ A = 900, and AB = AC
Therefore, ∠ B and ∠ C = ?
In Δ ABC
∠ A = 900 and
AB = AC
Therefore, ∠ B = ∠ C -------- (i)
(Angles opposite to equal sides of a triangle are equal)
Now, we know that, sum of all angles in a triangle = 1800
Therefore, ∠ A + ∠ B + ∠ C = 1800
⇒ 900 + ∠ B + ∠ C = 1800
[Because ∠ A = 90]
⇒ 900 + ∠ B + ∠ B = 1800
[Because, ∠ B = ∠ C from equation (i)]
⇒ 900 + 2 ∠ B = 1800
⇒ 2 ∠ B = 1800 – 900
⇒ 2 ∠ B = 900
Therefore, ∠ B = 900 / 2
⇒ ∠ B = 450
And ∠ C = 450
[Because ∠ B = ∠ C]
Thus, ∠ B = ∠ C = 45 Answer
Question (8) Show that the angles of an equilateral triangle are 600 each.
Solution
Let, Δ ABC is an equilateral triangle.
Thus, AB = BC = AC
To prove Each angle = 600
i.e. ∠ A = ∠ B = ∠ C = 600
Proof
We know that in an equilateral triangle all angles are equal
This means ∠ A = ∠ B = ∠ C
And the sum of all angles of a triangle = 1800
Thus, in Δ ABC
∠ A + ∠ B + ∠ C = 180
⇒ ∠ A + ∠ A + ∠ A = 1800
[Because ∠ A = ∠ B = ∠ C]
⇒ 3 ∠ A = 1800
Therefore, ∠ A = 1800 / 3
⇒ ∠ A = 600
Since, all angles are equal in a equilateral triangle, thus each angle of given triangle = 600 Proved
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