Question 1. In two right triangles, one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.
Solution:
We have two right triangles ABC and DEF in which one side and acute angle of one are equal to the corresponding side and angles of the other, i.e.,
∠A = ∠D
BC = EF
Prove: Δ ABC ≅ Δ DEF
Proof:
From Δ ABC and Δ DEF
∠ B = ∠ E = 90° …….(i)[Right triangle]
BC = EF …….(ii)[Given]
∠A = ∠D ……(iii)[Given]
By AAS congruence criterion,
Δ ABC ≅ Δ DEF
Hence, proved.
Question 2. If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.
Solution:
In ABC be a triangle
Given that AD is bisector of∠EAC and AD || BC.
From the above figure,
∠1 = ∠2 [AD is a bisector of ∠ EAC]
∠1 = ∠3 [Corresponding angles]
and
∠2 = ∠4 [alternative angle]
Also, we have ∠3 = ∠4
AB = AC
So the sides AB and AC are equal so
Δ ABC is an isosceles triangle.
Question 3. In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.
Solution:
Let us considered Δ ABC is isosceles triangle
In which AB = AC and ∠B = ∠C
Given: ∠A = 2(∠B + ∠C)
Find: The value of ∠A, ∠B, and ∠C
So, we have
∠A = 2(∠B + ∠C)
∠A = 2(∠B + ∠B) [∠B = ∠C because Δ ABC is isosceles triangle]
∠A = 2(2 ∠B)
∠A = 4(∠B)
As we know that sum of angles in an isosceles triangle = 180°
So, ∠A + ∠B + ∠C = 180°
Now put the value of ∠A and ∠C, we get
4 ∠B + ∠B + ∠B = 180°
6 ∠B =180°
∠B = 30°
Here, ∠B = ∠C
∠B = ∠C = 30°
And ∠A = 4 ∠B
∠A = 4 x 30° = 120°
Hence, the value of ∠A = 120°, ∠B = 30°, and ∠C = 30°.
Question 4. PQR is a triangle in which PQ = PR and is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.
Solution:
In ΔPQR is a triangle
It is given that PQ = PR, and S, a line is drawn parallel to QR and intersecting PR at T
So, ST || QR.
Prove: PS = PT
As we know that PQ = PR, so the given △PQR is an isosceles triangle.
Hence, ∠ PQR = ∠ PRQ
∠ PST = ∠ PQR and ∠ PTS = ∠ PRQ [Corresponding angles as ST parallel to QR]
∠ PQR = ∠ PRQ
∠ PST = ∠ PTS
So, In Δ PST,
∠ PST = ∠ PTS
Therefore, Δ PST is an isosceles triangle.
So, PS = PT.
Hence, proved.
Question 5. In an △ABC, it is given that AB = AC and the bisectors of B and C intersect at O. If M is a point on BO produced, prove that ∠MOC = ∠ABC.
Solution:
In △ABC,
It is given that AB = AC and the bisector of ∠B and ∠C intersect at O.
Prove: ∠MOC = ∠ABC
It is given that AB = AC
So the △ABC is an isosceles triangle
Hence
∠B = ∠C
∠ABC = ∠ACB
From the figure BO and CO are bisectors of ∠ABC and ∠ACB
So,
∠ABO = ∠OBC = ∠ACO = ∠OCB = 1/2 ∠ABC = 1/2 ∠ACB ………..(i)
In △OBC
∠BOC + ∠MOC = 180° ………(ii) [Straight line]
∠OBC + ∠OCB + ∠BOC = 180°[Sum of angles in an isosceles triangle = 180°]
From equation (ii)
∠OBC + ∠OCB + ∠BOC = ∠BOC + ∠MOC
∠OBC + ∠OCB = ∠MOC
2∠OBC = ∠MOC
From Equation(i)
2(1/2 ∠ABC) = ∠MOC
∠ABC = ∠MOC
Hence proved
Question 6. P is a point on the bisector of an angle ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.
Solution:
Given that P is a point on the bisector of ∠ABC, and PQ || AB.
Prove: △BPQ is an isosceles triangle
It is given that BP is bisector of ∠ABC
So, ∠ABP = ∠PBC ……….(i)
Also given that PQ || AB
So, ∠BPQ = ∠ABP ……….(ii) [alternative angles]
From equation (i) and (ii), we have
∠BPQ = ∠PBC
or ∠BPQ = ∠PBQ
In △BPQ
∠BPQ = ∠PBQ
Hence, proved △BPQ is an isosceles triangle.
Question 7. Prove that each angle of an equilateral triangle is 60°.
Solution:
Prove: Each angle of an equilateral triangle is 60°.
Let us considered we have an equilateral triangle △ABC
So, AB = BC = CA
Take AB = BC
So, ∠A = ∠C …….(i) [Because opposite angles to equal sides are equal]
Take BC = AC
∠B = ∠A ………(ii) [Because opposite angles to equal sides are equal]
From (i) and (ii), we get
∠A = ∠B = ∠C ………….(iii)
As we already know that the sum of angles in a triangle = 180°
So, ∠A + ∠B + ∠C = 180°
From equation(iii), we get
∠A + ∠A + ∠A = 180°
3∠A = 180°
∠A = 60°
So, ∠A = ∠B = ∠C = 60°
Hence Proved
Question 8. Angles ∠A, ∠B, ∠C of a triangle ABC are equal to each other. Prove that ABC is equilateral.
Solution:
Given that in △ABC
∠A =∠B = ∠C
Prove: △ABC is equilateral
In △ABC
As we already know that the sum of angles in a triangle = 180°
So, ∠A + ∠B + ∠C = 180°
Given that ∠A =∠B = ∠C
So,
∠A + ∠A + ∠A = 180°
3∠A = 180°
∠A = 60°
So, ∠A =∠B = ∠C = 60°
As we know that the angles of equilateral triangles are of 60°
Hence, proved that △ABC is equilateral.
Question 9. ABC is a triangle in which ∠B = 2 ∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove that ∠BAC = 72°.
Solution:
Given that in △ABC,
∠B = 2 ∠C, AD bisectors of∠BAC, and AB = CD.
Prove:∠BAC = 72°
Now, draw a line BP which is bisector of ∠ABC, and join PD.
Let us considered∠C = ∠ACB = y
∠B = ∠ABC = 2y
Let us considered ∠BAD = ∠DAC = x
∠BAC = 2x [AD is the bisector of ∠BAC]
In △BPC,
∠CBP = ∠PCB = y [BP is the bisector of ∠ABC]
PC = BP
In △ABP and △DCP,
∠ABP = ∠DCP = y
AB = DC [Given]
And PC = BP
So, by SAS congruence criterion,
△ABP ≅ △DCP
∠BAP = ∠CDP and AP = DP [C.P.C.T]
∠CDP = 2x
∠ADP = ∠DAP = x
In △ABD
∠ABD + ∠BAD + ∠ADB = 180°
∠ADB + ∠ADC = 180°
So,
∠ABD + ∠BAD + ∠ADB = ∠ADB + ∠ADC
2y + x = ∠ADP + ∠PDC
2y + x = x + 2x
2y = 2x
y = x
In △ABC,
As we already know that the sum of angles in a triangle = 180°
∠A + ∠B + ∠C = 180°
2x + 2y + y = 180° [∠A = 2x, ∠B = 2y, ∠C = y]
2(y) + 3y = 180° [x = y]
5y = 180°
y = 36°
∠A = ∠BAC = 2 × 36 = 72°
∠A = 72°
Hence proved
Question 10. ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
Solution:
Given that in ABC is a right-angled triangle
∠A = 90° and AB = AC
Find: ∠B and ∠C
In △ABC
AB = AC [Given]
∠B = ∠C
As we already know that the sum of angles in a triangle = 180°
So, ∠A + ∠B + ∠C = 180°
90° + ∠B + ∠B = 180°
2 ∠B = 180° – 90°
∠B = 45°
Hence, the value of ∠B = ∠C = 45°.