Prove theorem 2 of circle theory

Question

ScanSolve AI Tutor · Accepted Answer

The theorem you are referring to as "Theorem 2 of circle theory" is commonly known as the Angle at Centre Theorem.

Theorem Statement: The angle subtended by an arc at the centre of a circle is double the angle subtended by it at any point on the remaining part of the circle.

Let's prove this theorem.

Given: A circle with centre $O$ . $A$ , $B$ , and $C$ are three points on the circumference of the circle. Arc $AB$ subtends $\angle AOB$ at the centre and $\angle ACB$ at point $C$ on the remaining part of the circle.

To Prove: $\angle AOB = 2 \angle ACB$ .

We will consider three cases for the position of point $C$ relative to the arc $AB$ .

Case 1: The centre $O$ lies on one of the arms of $\angle ACB$ (e.g., on $AC$ ).

Draw a line segment from $C$ through $O$ to a point $D$ on the circumference (if $C, O, A$ are collinear, then $D$ is $A$ ).
In this specific case, $O$ lies on $AC$ .
Consider $\triangle BOC$ .
$OB = OC$ (radii of the same circle).
Therefore, $\triangle BOC$ is an isosceles triangle.
Angles opposite to equal sides in an isosceles triangle are equal.
$\angle OBC = \angle OCB$
The exterior angle of a triangle is equal to the sum of the two opposite interior angles.
For $\triangle BOC$ , $\angle AOB$ is an exterior angle.
$\angle AOB = \angle OBC + \angle OCB$
Substitute $\angle OBC = \angle OCB$ into the equation from Step 3.
$\angle AOB = \angle OCB + \angle OCB$ $\angle AOB = 2 \angle OCB$
Since $\angle OCB$ is the same as $\angle ACB$ in this case,
$\angle AOB = 2 \angle ACB$
Case 2: The centre $O$ lies inside $\angle ACB$ .
Draw a line segment from $C$ through $O$ to a point $D$ on the circumference.
This line divides $\angle ACB$ into two angles, $\angle ACD$ and $\angle BCD$ .
Applying Case 1 to arc $AD$ and angle $\angle ACD$ :
$\angle AOD = 2 \angle ACD$
Applying Case 1 to arc $BD$ and angle $\angle BCD$ :
$\angle BOD = 2 \angle BCD$
Adding these two equations:
$\angle AOD + \angle BOD = 2 \angle ACD + 2 \angle BCD$ $\angle AOB = 2 (\angle ACD + \angle BCD)$ $\angle AOB = 2 \angle ACB$
Case 3: The centre $O$ lies outside $\angle ACB$ .
Draw a line segment from $C$ through $O$ to a point $D$ on the circumference.
Applying Case 1 to arc $AD$ and angle $\angle ACD$ :
$\angle AOD = 2 \angle ACD$
Applying Case 1 to arc $BD$ and angle $\angle BCD$ :
$\angle BOD = 2 \angle BCD$
Subtracting the second equation from the first:
$\angle AOD - \angle BOD = 2 \angle ACD - 2 \angle BCD$ $\angle AOB = 2 (\angle ACD - \angle BCD)$ $\angle AOB = 2 \angle ACB$

Prove theorem 2 of circle theory

Answer

Step-by-step solution

Draw a line segment from $C$ through $O$ to a point $D$ on the circumference (if $C, O, A$ are collinear, then $D$ is $A$ ).

Angles opposite to equal sides in an isosceles triangle are equal.

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Substitute $\angle OBC = \angle OCB$ into the equation from Step 3.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Prove theorem 2 of circle theory

Answer

Step-by-step solution

Draw a line segment from CCC through OOO to a point DDD on the circumference (if C,O,AC, O, AC,O,A are collinear, then DDD is AAA).

Angles opposite to equal sides in an isosceles triangle are equal.

The exterior angle of a triangle is equal to the sum of the two opposite interior angles.

Substitute ∠OBC=∠OCB\angle OBC = \angle OCB∠OBC=∠OCB into the equation from Step 3.

Try one yourself

Still stuck on this one?

More Mathematics Questions

Draw a line segment from $C$ through $O$ to a point $D$ on the circumference (if $C, O, A$ are collinear, then $D$ is $A$ ).

Substitute $\angle OBC = \angle OCB$ into the equation from Step 3.