Theorems on Arc and Angle

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.

The following are circle theorems on arc and angle.
The Arc and Angle means the angle is subtended in the given arc.

Theorems on Arc and Angle

1) The angle subtended by an arc of a circle at the center is double the angle subtended by it at any point on the remaining part of the circle.
Given : Arc PQ of a circle C(O, r) and point r on the remaining part of the circle.
Prove that: ∠ POQ = 2∠PRQ
Construction: join RO and produce it to point M outside the circle.

Case 1: When arc PQ is a minor arc (figure (i))
Statements
Reasons
1) ∠POM = ∠OPM + ∠ORP 1) By exterior angle theorem (∠POM is exterior angle)
2) OP = OR 2) Radii of same circle
3) ∠OPR = ∠ORP 3) In a Δ two sides are equal then the angle opposite to them are also equal.
4) ∠POM = ∠ORP + ∠ORP 4) Substitution property. From (1)
5) ∠POM = 2∠ORP 5) Addition property.

Similarly, when ∠QOM is an exterior angle then ∠QOM = 2∠ORQ
∴ from above, ∠POQ = 2∠PRQ

Case 2: When arc PQ is a semicircle.[figure(ii)]
Statements
Reasons
1) ∠POM = ∠OPR + ∠ORP 1) Exterior angle theorem.
2) ∠POM = ∠ORP + ∠ORP 2) As,OP = OR = radius. ∴∠ORQ = ∠ORP
3) ∠POM = 2∠ORP 3) Substitution and addition property.
4) ∠QOM = ∠ORQ + ∠OQR 4) Exterior angle theorem.
5) ∠QOM = ∠ORQ + ∠ORQ 5) As, OQ =OR = radius, ∴ ∠ORQ = ∠OQR
6) ∠QOM = 2∠ORQ 6) Substitution and addition property.

∴ From above, ∠POQ = 2∠PRQ

Case 3: When arc PQ is a major arc.[figure(iii)]
Statements
Reasons
1) ∠POM = ∠OPR + ∠ ORP 1) By exterior angle theorem.
2) ∠POM = 2∠ORP 2) As, OP = OR = radius, ∴∠OPR = ∠ORP and by addition property
3) ∠ QOM = ∠ORQ + ∠OQR 3) By exterior angle theorem in ΔQOR
4) ∠QOM = 2∠ORQ 4) As, OP = OR = radius, ∴∠ORQ = ∠OQR and by addition property
5) ∠POM + ∠QOM = 2(∠ORP + ∠ORQ) 5) From (2) and (4)
6) Reflex ∠POQ = 2∠PRQ 6) Reflex of ∠POQ =∠POM + ∠QOM
2) The angle in a semicircle is a right angle.
Given : PQ is a diameter of the circle. ∠PRQ is an angle in a semicircle.
Prove: ∠PRQ = 90°

Statements
Reasons
1) ∠POQ = 2∠PRQ 1) Angle subtended by an arc of a circle at its center is twice the angle formed by the same arc.
2) 180 0 = 2∠PRQ 2) As POQ is a straight line
3) ∴ ∠PRQ = 90 0 3) Division property
So the angle in a semicircle is a right angle.
Theorems on Circle

• Theorems on Chord
• Theorems on Chord and Subtended Angle
• Theorems on Arc and Angle
• Theorems on Cyclic Quadrilateral

Home Page

Russia-Ukraine crisis update - 3rd Mar 2022

The UN General assembly voted at an emergency session to demand an immediate halt to Moscow's attack on Ukraine and withdrawal of Russian troops.