A) proving angle properties of circles 1
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This document summarizes how to prove that the angle subtended by an arc at the centre of a circle is twice that subtended at the circumference, even when a line cannot be drawn through the centre cutting both angles. It does this by extending one arc line to the centre, then using properties of isosceles triangles and angles on a straight line to show that the central angle is equal to twice the angle at the circumference.
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A) proving angle properties of circles 2
This document summarizes how to prove that the angle subtended by an arc at the centre of a circle is twice that subtended at the circumference, even when a line cannot be drawn through the centre cutting both angles. It does this by extending one arc line to the centre, then using properties of isosceles triangles and angles on a straight line to derive an equation that solves to the desired property.
A) proving angle properties of circles 2
This document proves the angle property that the angle subtended by an arc at the center of a circle is twice the angle subtended at the circumference. It provides an alternative proof for when a line cannot be drawn through the center and both angles. It does this by extending one arc line to the center, using properties of isosceles triangles and exterior angles to set up an equation relating the angles, and solving the equation to prove the property.
Proof of the property
The document proves that the angle at the center of a circle is twice the angle at the circumference. It draws a diameter from the center O through a point on the circle P. It defines the angles α and β and uses properties of circles, angles, and the diameter to show that the central angle AOB is equal to 2α, which is twice the angle at the circumference α.
Part A) Proof
Method One proves that the angle at the centre (Angle AOB) is twice the angle at circumference (Angle APB) by showing that triangles OBP and OAP are isosceles triangles, which allows calculating the angles.
Method Two extends a radius to form a diameter and shows that triangles OCA, OAP and OBP are isosceles triangles. It then uses properties of exterior angles to calculate that the angle at the centre (Angle AOB) is twice the angle at circumference (Angle APB). Both methods demonstrate the relationship between the angles at the centre and circumference of a circle.
Math Proof: Angle at circumference is twice angle at centre
The document proves that the angle at the center of a circle is twice the angle at the circumference through a series of logical steps and equations. It first draws a diagram with lines and identifies equal radii and isosceles triangles. It then establishes two initial equations relating angles and substitutes one into the other to derive a third equation. Further steps substitute known angle relationships regarding triangles to eliminate variables until the final equation shows the relationship between the center and circumference angles.
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The document proves that in a circle with lines OA, OB, and OP subtending from the center, angle AOB is twice angle APB. Angle OBP is shown to equal the sum of angles OPA and APB. The exterior angle theorem is then applied to show angle ACB is the sum of angles APB and OBP, and angle AOC is equal to angle ACB minus angle OPA, demonstrating angle AOB is twice angle APB.
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This document summarizes how to prove that the angle subtended by an arc at the centre of a circle is twice that subtended at the circumference, even when a line cannot be drawn through the centre cutting both angles. It does this by extending one arc line to the centre, then using properties of isosceles triangles and angles on a straight line to derive an equation that solves to the desired property.
A) proving angle properties of circles 2
This document proves the angle property that the angle subtended by an arc at the center of a circle is twice the angle subtended at the circumference. It provides an alternative proof for when a line cannot be drawn through the center and both angles. It does this by extending one arc line to the center, using properties of isosceles triangles and exterior angles to set up an equation relating the angles, and solving the equation to prove the property.
Proof of the property
The document proves that the angle at the center of a circle is twice the angle at the circumference. It draws a diameter from the center O through a point on the circle P. It defines the angles α and β and uses properties of circles, angles, and the diameter to show that the central angle AOB is equal to 2α, which is twice the angle at the circumference α.
Part A) Proof
Method One proves that the angle at the centre (Angle AOB) is twice the angle at circumference (Angle APB) by showing that triangles OBP and OAP are isosceles triangles, which allows calculating the angles.
Method Two extends a radius to form a diameter and shows that triangles OCA, OAP and OBP are isosceles triangles. It then uses properties of exterior angles to calculate that the angle at the centre (Angle AOB) is twice the angle at circumference (Angle APB). Both methods demonstrate the relationship between the angles at the centre and circumference of a circle.
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The document proves that in a circle with lines OA, OB, and OP subtending from the center, angle AOB is twice angle APB. Angle OBP is shown to equal the sum of angles OPA and APB. The exterior angle theorem is then applied to show angle ACB is the sum of angles APB and OBP, and angle AOC is equal to angle ACB minus angle OPA, demonstrating angle AOB is twice angle APB.
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A) proving angle properties of circles 1
- 1. (a) Proving Angle Properties of Circles
- 2. A B P O C that the angle subtended by an arc at the centre of a circle is twice that subtended at the circumference of the same circle . ∠ AOB = 2 x ∠APB e.g. If ∠AOB = 50°, then according to the property, ∠APB = 25°. In Geometry, there is a property which states However for this proof, it is assumed that we are always able to draw a straight line (i.e. POC) which passes through the centre O , and cuts both ∠AOB and ∠APB .
- 3. A B P O But what you were given this figure where it is impossible to draw a straight line that passes through the centre O and still can cut both ∠AOB and ∠APB ? How do you proof it then?
- 4. Let us show you how to :D
- 5. A B P O 1 Extend line AO 2 Draw a line from O to P C
- 6. A B P O ∠ AOB + ∠BOP + ∠POC = 180° (adj. angles on a straight line) C ∠ BOP = 180 ° – 2 x ∠OPB ∠ OBP = ∠OPB (isosceles triangle) P O
- 7. A B P O C ∠ AOB + 180 ° – 2 x ∠OPB + ∠POC = 180 ° ∠ OPB = ∠OPA + ∠APB B P O A ∠ AOB – 2( ∠OPA + ∠APB ) + ∠POC = O ∠ AOB – 2 ∠OPA – 2 ∠APB + ∠POC = O
- 8. A B P O C ∠ POC = 2 ∠OPA [ ∠OPA = ∠OAP (isosceles triangle), ∠ POC = ∠OAP + ∠OPA (external angles)] B P O C A ∠ AOB - 2 ∠OPA - 2 ∠APB + ∠POC = O
- 9. A B P O C Therefore… ∠ AOB = 2 ∠APB PROVEN !! :D
- 10. The end :D