Sponsor Area
∠FOB + ∠BOE = 180° ...(1)
| Linear Pair Axiom
∠FOA + ∠AOE = 180° ...(2)
| Linear Pair Axiom
From (1) and (2),
∠FOB + ∠BOE = ∠FOA + ∠AOE ...(3)
But ∠BOE = ∠AOE
| ∵ Ray OE bisects ∠AOB
∴ From (3),
⇒ ∠FOB = ∠FOA.
Rays OA, OB. OC, OD and OE have the common initial point O. Show that ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.
Construction. Draw a ray OP opposite to ray OA.
Proof. ∠AOB + ∠BOC + ∠COP = 180° ...(1)
| ∵ A straight angle = 180°
∠POD + ∠DOE + ∠EOA = 180° ...(2)
| ∵ A straight angle = 180°
Adding (1) and (2), we get
∠AOB + ∠BOC + (∠COP + ∠POD) + ∠DOE + ∠EOA = 180° + 180° = 360°
⇒ ∠AOB + ∠BOC + ∠COD + ∠DOE + ∠EOA = 360°.
This leads to two pairs of vertically opposite angles, namely,
(i) ∠AOC and ∠BOD
(ii) ∠AOD and ∠BOC
We are to prove that
(i) ∠AOC = ∠BOD
and (ii) ∠AOD = ∠BOC
∵ Ray OA stands on line CD Therefore,
∠AOC + ∠AOD = 180° ...(1)
| Linear Pair Axiom
∵ Ray OD stands on line AB Therefore,
∠AOD + ∠BOD = 180° ...(2)
| Linear Pair Axiom
From (1) and (2),
∠AOC + ∠AOD = ∠AOD + ∠BOD
⇒ ∠AOC = ∠BOD
Similarly, we can prove that
∠AOD = ∠BOC
Given: OP bisects ∠AOC, OQ bisects ∠BOC and OP ⊥ OQ.
To Prove: The points A, O and B are collinear.
Proof: ∵ OP bisects ∠AOC
∴ ∠AOP = ∠COP ...(1)
∵ OQ bisects ∠BOC
∠BOQ = ∠COQ ...(2)
Now, ∠AOB
= ∠AOP + ∠COP + ∠COQ + ∠BOQ
= ∠COP + ∠COP + ∠COQ + ∠COQ
| From (1) and (2)
= 2(∠COP + ∠COQ)
= 2 ∠POQ
= 2(90°) | ∵ OP ⊥ OQ
= 180°
∴ The points A, O and B are collinear.
| By converse of Linear Pair Axiom
∵ Sum of all the angle round a point is equal to 360°.
∴ y + (3x - 15) + (y + 5) + 2y + (4y + 10) + x = 360°
⇒ 4x + 8y = 360°
⇒ x + 2y = 90°
⇒ x + 2(20°) = 90°
⇒ x + 40° = 90°
⇒ x = 50°
Now, y + 3x - 15 + y + 5 = 3x + 2y - 10
= 3(50°) + 2(20°) - 10
= 150° + 40° - 10°
= 180°
∴ AOB is a straight line.
∵ BOA is a line
(∠AOC + ∠BOE) + ∠EOC = 180°
⇒ 70° + EOC = 180°
⇒ ∠EOC = 110°
⇒ Reflex ∠EOC = 360° - 110° = 250°
Solution not provided.
Ans. ∠POR = 75°, ∠ROQ = 105°, ∠POS = 105°, ∠SOQ = 75°
Solution not provided.
Ans. 90°
Solution not provided.
Ans. 45°
Solution not provided.
Ans. 130°, 50°
Solution not provided.
Ans. 16
Solution not provided.
Ans. 80°, 100°
Solution not provided.
Ans. 80°
An angle is equal to five times of its complement. Find the measure of the angle.
Solution not provided.
Ans. 75°
Solution not provided.
Ans. 140°, 40°, 140°,
Sponsor Area
Solution not provided.
Ans. Yes
Solution not provided.
Ans. 70°
∵ Lines AB and CD intersect at O
∴ ∠AOC = ∠BOD
| Vertically Opposite Angles
But ∠BOD = 40° ...(1) | Given
∴ ∠AOC = 40° ...(2)
Now, ∠AOC + ∠BOE = 70°
⇒ 40° + ∠BOE = 70° | Using (2)
⇒ ∠BOE = 70° - 40°
⇒ ∠BOE = 30°
Again,
Reflex ∠COE
= ∠COD + ∠BOD + ∠BOE
= ∠COD + 40° + 30°
| Using (1) and (2)
= 180° + 40° + 30°
| ∵ Ray OA stands on line CD
|∴ ∠AOC + ∠AOD = 180° (Linear Pair Axiom) ⇒ ∠COD = 180°
= 250°.
Ray OP stands on line Xy
a + b = 900 .....(1)
a = 2k
∵ Ray OX stands on line MN
∴ ∠XOM + ∠XON = 180°
| Linear Pair Axiom
⇒ b + c = 180°
⇒ 54° + c = 180° | Using (2)
⇒ c = 180° - 54°
⇒ c = 126°.
∵ Ray QP stands on line ST
∴ ∠PQS + ∠PQR = 180° ...(1)
| Linear Pair Axiom
∵ Ray RP stands on line ST
∴ ∠PRQ + ∠PRT = 180° ...(2)
| Linear Pair Axiom
From (1) and (2), we obtain
∠PQS + ∠PQR = ∠PRQ + ∠PRT
⇒ ∠PQS = ∠PRT.
| ∵ ∠PQR = ∠PRQ (Given)
x + y = w + z ...(1) | Given
∵ The sum of all the angles round a point is equal to 360°.
x + y + w + z = 3600 
x + y + x + y = 3600 | Using (1)
2(x + y) = 3600
x + y = 1800
∴ AOB is a line.
| If the sum of two adjacent angles is 180°, then the non-common arms of the angles form a line.
∵ Ray OR is perpendicular to line PQ.
∴ ∠QOR = ∠POR = 90° ...(1)
∠QOS = ∠QOR + ∠ROS ...(2)
∠POS = ∠POR - ∠ROS ...(3)
From (2) and (3),
∴ ∠QOS - ∠POS = (∠QOR - ∠POR) + 2∠ROS = 2∠ROS | Using (1)
∵ Ray YZ stands on line PX
∴ ∠XYZ+ ∠ZYP = 180°
| Linear Pair Axiom
⇒ 64° + ∠ZYP = 180°
| ∵ ∠XYZ = 64° (Given)
⇒ ∠ZYP = 180° - 64°
⇒ ∠ZYP = 116° ...(1)
Ray YQ bisects 
| Using (1)
= 580 ...(2)
The sum of all the angles round apoint is equal to 3600
Again, 
= 640 + 580
(Given) and 
[ From (2) ]
= 1220.
∵ XOY is a line
∴ b + a + 90° = 180°
a + b = 900
2 + 3 = 5
∵ Ray AE stands on line GH ∴AEG + ∠AEH = 180°
| Linear Pair Axiom
⇒ 50° + x = 180°
⇒ x = 180° - 50° = 130° ...(1)
y = 130° ...(2)
| Vertically Opposite Angles
From (1) and (2), we conclude that
x = y
But these are alternate interior angles and they are equal.
So, we can say that AB || CD.
∵ AB || CD
and CD || EF
∴ AB || EF
| Lines parallel to the same line are parallel to each other
∴ x = z ...(1)
| Alternate Interior Angles
x + y = 180° ...(2)
| Consecutive interior angles on the same side of the transversal GH to parallel lines AB and CD
From (1) and (2),
z + y = 180°
y : z = 3 : 7
Sum of the ratios = 3 + 7 = 10
(i) ∠AGE = ∠GED = 126°
| Alternate Interior Angles
(ii) ∠GED = 126°
⇒ ∠GEF + ∠FED = 126°
⇒ ∠GEF + 90° = 126°
| ∵ EF ⊥ CD ∴ ∠FED = 90°
⇒ ∠GEF = 126° - 90° = 36°
(iii) ∠GEC + ∠GEF + ∠FED = 180°
| ∵ CD is a line
⇒ ∠GEC + 36° + 90° = 180°
⇒ ∠GEC + 126°= 180°
⇒ ∠GEC = 180° - 126° = 54°
Now, ∠FGE = ∠GEC = 54°
| Alternate Interior Angles
In figure, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.
[Hint. Draw a line parallel to ST through point R.]
∠RST + ∠SRU = 180°
| Sum of the consecutive interior angles on the same side of the transversal is 180°
⇒ 130° + ∠SRU = 180°
⇒ ∠SRU = 180° - 130° = 50° ...(1)
∠QRU = ∠PQR = 110°
| Alternate Interior Angles
⇒ ∠QRS + ∠SRU = 110°
⇒ ∠QRS + 50° = 110° | Using (1)
⇒ ∠QRS = 110° - 50° = 60°.
x = ∠APQ = 50°
| Alternate Interior Angles
∠APQ + y = ∠PRD = 127°
| Alternate Interior Angles
50° + y = 127°
y = 127° - 50° = 77°
Proof: ∵ BL ⊥ PQ, CM ⊥ RS and PQ || RS ∴ BL || CM
∠LBC = ∠MCB ...(1)
| Alternate Interior Angles
∠ABL = ∠LBC ...(2)
| ∵ Angle of incidence = Angle of reflection
∠MCB = ∠MCD ...(3)
| ∵ Angle of incidence = Angle of reflection
From (1), (2) and (3), we get
∠ABL = ∠MCD ...(4)
Adding (1) and (4), we get
∠LBC + ∠ABL = ∠MCB + ∠MCD
⇒ ∠ABC = ∠BCD
But these are alternate interior angles and they are equal.
So, AB || CD.
In figure, if x = y and a = b, prove that r || n.
r and m are two lines and a transversal p intersects them such that
x = y
But these angles form a pair of equal corresponding angles
∴ r || m ...(1)
Again, m and n are two lines and a transversal q intersects them such that
a = b
But these angles form a pair of equal corresponding angles
∴ m || n ...(2)
From (1) and (2), we have r || n.
Then,
∠1 = 90°
∠2 = 90°
∴ ∠1 = ∠2
But these angles form a pair of equal corresponding angles
∴ m || n
In figure, EF is a transversal to two parallel lines AB and CD, GM and HL are the bisectors of the corresponding angles EGB and EHD. Prove that GM || HL.
[Hint. First prove that ∠EGM = ∠GHL]
AB || CD and a transversal EF intersects them
∴ ∠EGB = ∠GHD
| Corresponding Angles
⇒ 2 ∠EGM = 2 ∠GHL
| ∵ GM and HL are the bisectors of ∠EGB and ∠EHD respectively.
⇒ ∠EGM = ∠GHL
But these angles form a pair of equal corresponding angles for lines GM and HL and transversal EF.
∴ GM || HL.
Construction: Through O draw OE || AB || CD
Now, y = ∠FOG
= ∠FOE + ∠GOE
= ∠CFO + ∠AGO
∠FOE = ∠CFO
(Alternate Interior Angles)
∠GOE = ∠AGO (Alternate Interior Angles)
= 45° + 40°
= 85°
We see that
142° + 38° = 180°, i.e., the sum of the consecutive interior angles on the same side of
the transversal m is 180°.
So, by converse of consecutive interior angles axiom,
l || m
In figure, PQ || RS and T is any point as shown in the figure then show that
∠PQT + ∠QTS + ∠RST = 360°.
Given: PQ || RS and T is any point.
To Prove: ∠PQT + ∠QTS + ∠RST = 360°
Construction: Through T, draw TU || PQ || RS
Proof: ∵ PQ || UT | By construction and a transversal QT intersects then
∴ ∠PQT + ∠QTU = 180° ...(1)
| Sum of consecutive interior angles on the same sides of a transversal is 180°
∵ UT || RS | By construction and a transversal TS intersects them
∴ ∠UTS + ∠RST = 180° ...(2)
| Sum of consecutive interior angles on the same side of a transversal is 180°
Adding (1) and (2), we get
∠PQT + (∠QTU + ∠UTS) + ∠RST = 360°
⇒ ∠PQT + ∠QTS + ∠RST = 360°
Sponsor Area
Given: l, m, n are three lines such that l || m and n ⊥ l.
To Prove: n ⊥ m
Proof: ∵ l || m and n is a transversal
∵ ∠1 = ∠2 | Corresponding angles
But ∠1 = 90° | ∵ n ⊥ l (given)
∴ ∠2 = 90°
⇒ n ⊥ m
Given: l || m
To Prove: ∠1 + ∠2 - ∠3 = 180°
Construction: Through C, draw CF || l || m
Proof: ∵ l || CF by construction and a transversal BC intersects them
∴ ∠1 + ∠FCB = 180°
| ∵ Sum of consecutive interior angles on the same side of a transversal is 180°
⇒ ∠1 + ∠2 - ∠FCD = 180° ...(1)
But ∠FCD = ∠3 ...(2)
| Alternate interior angles
∴ From (1) and (2),
∠1 + ∠2 - ∠3 = 180°
∵ CD || EF and a transversal DE intersects them
∴ y + 40° = 180°
| Sum of consecutive interior on the same side of a transversal is 180°
⇒ y = 180° - 40° = 140°
∵ AB || CD and a transversal BD intersects them
x = y | Corresponding angles
⇒ x = 140°
∵ EA ⊥ AB and AB || EF
∵ EA ⊥ EF
| If a line is perpendicular to a line, then it is perpendicular to the parallel line also
⇒ ∠AEF = 90°
⇒ z + 40° = 90°
⇒ z = 50°
∠1 = 4x - 15 | Corresponding angles
2x = ∠1 = 4x - 15
| Corresponding angles
Given: Two parallel lines AB and CD are intersected by a transversal EF in points G and H respectively. The bisectors of two pairs of interior angles intersect in L and M.
To Prove: GLHM is a rectangle.
Proof: ∵ AB || CD and a transversal EF intersects them
| Alternate interior angles
| Halves of equals are equal
But these form a pair of equal alternate interior angles
∴ GM || HL ...(1)
Similarly, we can show that
HM || GL ...(2)
In view of (1) and (2),
GLHM is a parallelogram
| A quadrilateral is a parallelogram if its both the pairs of opposite sides are parallel
Now, since the sum the consecutive interior angles on the same side of a transversal is 180°
| Halves of equals are equal
⇒ ∠3 + ∠2 = 90° ...(3)
In ∆GHL,
∠3 + ∠2 + ∠GLH = 180°
| Angle sum property of a triangle
⇒ 90° + ∠GLH = 180°
| From (3)
∠GLH = 90°
⇒ GLHM is a rectangle
| A parallelogram with one of its angles of measure 90° is a rectangle.
Solution not provided.
Ans. 850
Solution not provided.
Ans. x = y = 125°, z = 35°.
Solution not provided.
Ans. 125°, 125°, 35°
Solution not provided.
Ans. 80°, 100°
Solution not provided.
Ans. 270°
Solution not provided.
Ans. 50°, 77°
Solution not provided.
Ans. 80°
∵ TR is a line
∴ ∠PQT + ∠PQR = 180°
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 180° - 110° = 70° ...(1)
∵ QS is a line
∴ ∠SPR + ∠QPR = 180°
⇒ 135° + ∠QPR = 180°
⇒ ∠QPR = 180° - 135° = 45° ...(2)
In ∆PQR,
∠PQR + ∠QPR + ∠PRQ = 180°
| ∵ The sum of all the angles of a triangle is 180°
⇒ 70° + 45° + ∠PRQ = 180°
| Using (1) and (2)
⇒ 115° + ∠PRQ = 180°
⇒ ∠PRQ = 180° - 115° = 65°.
In ∆XYZ,
∠XYZ + ∠YZX + ∠ZXY = 180°
| ∵ The sum of all the angles of a triangle is 180°
⇒ 54° + ∠YZX + 62° = 180°
⇒ 116° + ∠YZX = 180°
⇒ ∠YZX = 180° - 116° = 64° ...(1)
∵ YO is the bisector of ∠XYZ
In ∆OYZ,
∠OYZ + ∠OZY + ∠YOZ = 180°
|∵ The sum of all the angles of a triangle is 180°
⇒ 27° + 32° + ∠YOZ = 180°
| Using (2) and (3)
⇒ 59° + ∠YOZ = 180°
⇒ ∠YOZ = 180° - 59° = 121°.
∠DEC = ∠BAC = 35° ...(1)
| Alternate Interior Angles
∠CDE = 53° ...(2) | Given
In ∆CDE,
∠CDE + ∠DEC + ∠DCE = 180°
| ∵ The sum of all the angles of a triangle is 180°.
⇒ 53°+ 35° + ∠DCE = 180°
| Using (1) and (2)
⇒ 88° + ∠DCE = 180°
⇒ ∠DCE = 180° - 88° = 92°.
In ∆PRT,
∠PTR + ∠PRT + ∠RPT = 180°
| ∵ The sum of all the angles of a triangle is 180°
⇒ ∠PTR + 40°+ 95° = 180°
⇒ ∠PTR + 135° = 180°
∠ ∠PTR = 45°
⇒ ∠QTS = ∠PTR = 45°
| Vertically Opposite Angles
In ∆TSQ,
∠QTS + ∠TSQ + ∠SQT = 180°
| ⇒ The sum of all the angles of a triangle is 180°
⇒ 45° + 75° + ∠SQT = 180°
⇒ 120° + ∠SQT = 180°
⇒ ∠SQT = 180° - 120° = 60°.
∠QRT = ∠RQS + ∠QSR
| ∵ The exterior angle is equal to the sum of the two interior opposite angles. ⇒ 65°
= 28° + ∠QSR
⇒ ∠QSR = 65° - 28° = 37°
∵ PQ ≠ SP
∴ ∠QPS = 90°
∵ PQ || SR
∴ ∠QPS + ∠PSR = 180°
| ∵ The sum of consecutive interior angles on the same side of the transversal is 180°
⇒ 90° + ∠PSR = 180°
⇒ ∠PSR = 180° - 90° = 90°
⇒ ∠PSQ + ∠QSR = 90°
⇒ y + 37° = 90°
⇒ y = 90° - 37° = 53°
In ∆PQS,
∠PQS + ∠QSP + ∠QPS = 180°
| ∵ The sum of all the angles of a triangle is 180°
⇒ x + y + 90° = 180°
⇒ x + 53° + 90° = 180°
⇒ x + 143° = 180°
⇒ x = 180° - 143° = 37°.
∵ ∠TRS is an exterior angle of ∆TQR ∴ ∠TRS = ∠TQR + ∠QTR ...(1)
| ∵ The exterior angle is equal to sum of its two interior opposite angles
∵ ∠ PRS is an exterior angle of ∆PQR
∴ ∠PRS = ∠PQR + ∠QPR ...(2)
| ∵ The exterior angle is equal to the sum of its two interior opposite angles
⇒ 2 ∠TRS = 2∠TQR + ∠QPR
| ∵ QT is the bisector of ∠PQR and RT is the bisector of ∠PRS
⇒ 2(∠TRS - ∠TQR) = ∠QPR ...(3)
From (1),
∠TRS - ∠TQR = ∠QTR ...(4)
From (3) and (4), we obtain
Let in ∆ABC,
∠A = ∠B + ∠C ...(1)
We know that,
∠A + ∠B + ∠C = 180°
| ∵ The sum of the three angles of a triangle is 180°
⇒ ∠A + ∠A = 180° | From (1)
⇒ 2 ∠A = 180°
Hence triangle ABC is right angled triangle.
In ∆OBC, ∠BOC + ∠OBC + ∠OCB = 180°
| ∵ The sum of the three angles of a ∆ is 180°
| From (1) and (2)
In ∠ABC, ∠A + ∠B + ∠C = 180°
| ∵ The sum of the three angles of a triangle is 180°
From (3) and (4), we have
The side EF, FD and DE of a triangle DEF are produced in order forming three exterior angles DFP, EDQ and FER respectively. Prove that
∠DFP + ∠EDQ + ∠FER = 360°.
OR
Prove that the sum of the exterior angles of a triangl is 360°.
∠DFP = ∠D + ∠E ...(1)
| Exterior Angle Theorem
∠EDQ = ∠E + ∠F ...(2)
| Exterior Angle Theorem
∠FER = ∠F + ∠D ...(3)
| Exterior Angle Theorem
Adding (1), (2) and (3), we get
∠DFP + ∠EDQ + ∠FER = 2(∠D + ∠E + ∠F)
But ∠D + ∠E + ∠F = 180°
| ∵ The sum of the three angles of a triangle is 180°
∴ ∠DFP + ∠EDQ + ∠FER = 2(180°) = 360°.
Given: AD and CE are the angle bisectors of ∠A and ∠C respectively. ∠ABC = 90°
To Determine: ∠AOC
Determination: In ∆ABC,
∠A + ∠B + ∠C = 180°
| Angles sum property of a triangle
⇒ ∠A + 90° + ∠C = 180°
⇒ ∠A + ∠C = 90° ...(1)
In ∆AOC,
∠OAC + ∠OCA + ∠AOC = 180°
| Angle sum property of a triangle
Construction: Join BD and extend upto E.
x = ∠ADC
= ∠ADE + ∠CDE
= (∠DAB + ∠ABD) + (∠DBC + ∠DCB)
| Exterior angle theorem
= 35° + (∠ABD + ∠DBC) + 50°
= 35° + ∠ABC + 50°
= 35° + 45° + 50°
= 130°
In ∆QTR,
∠QTR + ∠r + 40° = 180°
| ∠ Angle sum property of a triangle
⇒ 90° + ∠x + 40° = 180°
⇒ ∠r + 130° = 180°
⇒ ∠x = 180° - 130° = 50°
∠x = 50°
In ∠PSR,
y° = x° + 30°
| Exterior angle theorem
= 50° + 30°
= 80°
⇒ y = 80°
In ∆OAD,
∠OAD + ∠ODA + ∠AOD = 180°
| Angle sum property of a triangle
⇒ 80° + 50° + ∠AOD = 180°
⇒ 130° + ∠AOD = 180°
⇒ ∠AOD = 180° - 130° = 50°
⇒ ∠BOC = 50°
| ∵ ∠AOD = ∠BOC (Vertically opposite angles)
In ∆OBC,
∠BOC + ∠OCB + ∠OBC = 180°
| Angle sum property of a triangle ⇒ 50° + 40° + ∠OBC = 180°
⇒ 90° + ∠OBC = 180°
∠ ∠OBC = 180° - 90° = 90°
Given: AP and DP are bisectors of two adjacent angles A and D of a quadrilateral ABCD.
To Prove: 2 ∠APD = ∠B + ∠C
Proof: We know that the sum of all the angles of a quadrilateral is 360°. So,
∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠A + ∠D = 360° - (∠B + ∠C) ...(1)
Now, in ∆PAD,
∠APD + ∠PAD + ∠PDA = 180°
| Angle sum property of a triangle
∵ AP and DP are the bisectors of two adjacent angles A and D of quadrilateral ABCD
⇒ 2 ∠APD + ∠A + ∠D = 360°
⇒ 2 ∠APD = 360° - (∠A + ∠D)
⇒ 2∠APD = ∠B + ∠C
Construction: Extend ED, AB and CB.
∠BHF = ∠CBA
| Corresponding angles
∵ AB || FE and BH intersects them
= 75° ...(1)
∠BHF + ∠BHD = 180° | Linear pair axiom
⇒ 75° + ∠BHD = 180°
⇒ ∠BHD = 105° ...(2)
∠CDH + ∠CDE = 180° | Linear pair axiom
⇒ ∠CDH + 145° = 180°
⇒ ∠CDH = 35° ...(3)
In ∆CHD,
∠CDH + ∠CHD + ∠BCD = 180°
| Angle sum property of a triangle
⇒ 35° + 105° + ∠BCD = 180°
⇒ ∠BCD = 40°
Given: The side BC of ∆ABC is produced to D. The bisector of ∠A meets BC in L.
To Prove: ∠ABC + ∠ACD = 2 ∠ALC.
Proof: ∠ABC + ∠ACD
= ∠ABC + (∠ABC + ∠BAC)
| Exterior angle theorem
= 2 ∠ABC + ∠BAC
= 2 ∠ABC + 2 ∠BAL
| ∵ AL is the bisector of ∠A
= 2 (∠ABC + ∠BAL)
= 2 ∠ALC | Exterior angle theorem
Given: A hexagon ABCDEF
To Prove:
∠ A + ∠B + ∠C + ∠D + ∠E + ∠F = 720°
Construction: Join AD, BE and FC so as to intersect at O.
Proof: In ∆OAB,
∠1 + ∠7 + ∠9 = 180° ...(1)
| Angle sum property of a triangle
In ∆AOBC,
∠2 + ∠10 + ∠11 = 180° ...(2)
| Angle sum property of a triangle
In ∆OCD,
∠3 + ∠12 + ∠13 = 180° ...(3)
| Angle sum property of a triangle
In ∆CDE,
∠4 + ∠14 + ∠15 = 180° ...(4)
| Angle sum property of a triangle
In ∆OEF,
∠5 + ∠16 + ∠17 = 180° ...(5)
| Angle sum property of a triangle
In ∆OFA,
∠6 + ∠18 + ∠8 = 180° ...(6)
| Angle sum property of a triangle
Adding (1), (2), (3), (4), (5) and (6), we get
(∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6)
+ (∠7 + ∠8) + (∠9 + ∠10)
+ (∠11 + ∠12) + (∠13 + ∠14)
+ (∠15 + ∠16) + (∠17 + ∠18)
=1080°
⇒ 360° + ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 1080°
| ∵ Sum of all the angles round a point is equal to 360°
⇒ ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 1080° - 360° = 720°
Given: PS is the bisector of ∠PQR and PT⊥QR.
To Prove: 
Proof: ∵ PS is the bisector of ∠QPR
∴ ∠QPS = ∠RPS
⇒ ∠1 + ∠TPS = ∠2 ...(1)
In ∆PQT,
∠PTQ = 90° | Given
∴ ∠1 + ∠Q = 90°
| Angle sum property of a triangle
⇒ ∠Q = 90° - ∠1 ...(2)
In ∆PRT,
∠PTR = 90° | Given
∴ ∠R + ∠TPR = 90°
| Angle sum property of a triangle
⇒ ∠R + (∠TPS + ∠2) = 90° ...(3)
From (2) and (3),
∠Q = ∠R + (∠TPS + ∠2) - ∠1
⇒ ∠Q - ∠R = ∠TPS + (∠2 - ∠1)
⇒ ∠Q - ∠R = ∠TPS + ∠TPS | From (1) ⇒ ∠Q - ∠R = 2 ∠TPS
Given: ABC is a triangle right angled at A. AL is drawn perpendicular to BC.
To Prove: ∠BAL = ∠ACB
Proof: In triangle ALB,
∠ALB + ∠BAL + ∠ABL = 180°
| Angle sum property of a triangle
⇒ 90° + ∠BAL + ∠ABC = 180°
⇒ ∠BAL + ∠ABC = 90° ...(1)
In triangle ABC,
∠BAC + ∠ACB + ∠ABC = 180°
| Angle sum property of a triangle
⇒ 90° + ∠ACB + ∠ABC = 180°
⇒ ∠ACB + ∠ABC = 90° ...(2)
From (1) and (2),
∠BAL + ∠ABC = ∠ACB + ∠ABC
⇒ ∠BAL = ∠ACB
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Given: ABCD is a quadrilateral
To Prove: ∠A + ∠B + ∠C + ∠D = 360°
Construction: Join AC.
Proof: In ∆ABC,
∠1 + ∠B + ∠3 = 180° ...(1)
| Angle sum property of a triangle
In ∆ADC,
∠2 + ∠D + ∠4 = 180° ...(2)
| Angle sum property of a triangle
Adding (1) and (2), we get
(∠1 + ∠2) + ∠B + (∠3 + ∠4) + ∠D = 360°
∠ ∠A + ∠B + ∠C + ∠D = 360°
Given: The side BC of a ∆ABC is produced to D. The bisector of ∠BAC intersects the side BC at E.
To Prove: ∠ABC + ∠ACD = 2 ∠AEC.
Proof: In ∆ABE,
∠AEC = ∠ABC + ∠BAE
| Exterior angle theorem
= ∠ABC + ∠CAE ...(1)
| ∠BAE = ∠CAE (∵ AE bisects ∠BAC)
In ∆AEC,
∠ACD = ∠AEC + ∠CAE
| Exterior angle theorem
⇒ ∠CAE = ∠ACD - ∠AEC ...(2)
From (1) and (2),
∠AEC = ∠ABC + (∠ACD - ∠AEC)
⇒ 2 ∠AEC = ∠ABC + ∠ACD
Given: The sides BA and DC of a quadrilateral ABCD are produced.
To Prove: ∠x + ∠y = ∠a + ∠b
Construction: Join BD
Proof: In ∆BCD,
∠a = ∠2 + ∠4 ...(1)
| Exterior angle theorem
In ∆ADB,
∠b = ∠1 + ∠3 ...(2)
| Exterior angle theorem
Adding (1) and (2), we get
∠a + ∠b = (∠1 + ∠2) + (∠3 + ∠4)
⇒ ∠a + ∠b - ∠x + ∠y
then find the value of z.
∵ x, y and z are the degree measures of three angles of a triangle
∴ x + y + z = 180°
| Angle sum property of a triangle
| Dividing throughout by 2
Solution not provided.
Ans. 20°
Solution not provided.
Ans. 50°, 30°, 100°,
Solution not provided.
Ans. 41°, 116°
Solution not provided.
Ans. 30°, 70°, 110°
Solution not provided.
Ans. 20°, 130°
Solution not provided.
Ans. 80°, 60°
Solution not provided.
Ans. 80°, 60°, 40°
Solution not provided.
Ans. 55°, 25°
Solution not provided.
Ans. 30°, 60°. 90°
Solution not provided.
Ans. 105°, 75°
Solution not provided.
Ans. 95°
Solution not provided.
Ans. 130°
Solution not provided.
Ans. 65°
Solution not provided.
Ans. 85°, 95°
A.
measures between 0° and 90°B.
is exactly equal to 90°B.
is greater than 90° but less than 180°D.
is exactly equal to 180°B.
a straight angleA.
is greater than 180° but less than 360°B.
complementary anglesA.
supplementary anglesSponsor Area
B.
two right anglesD.
Interior angles on the same side of the transversal are complementary.D.
Two acute anglesC.
Interior opposite angles
Solution not provided.
Ans. 140°, 140°, 140°
Solution not provided.
Ans. AOC is a straight line but not BOD
Solution not provided.
Ans. 30°, 70°, 110°
Solution not provided.
Ans. 90° 13. 110°
Solution not provided.
Ans. 15°
Solution not provided.
Ans. 30°, 30°
Sponsor Area
Sponsor Area
