Inscribed angle

In geometry, an inscribed angle is formed when two secant lines of a circle (or, in a degenerate case, when one secant line and one tangent line of that circle) intersect on the circle.

Typically, it is easiest to think of an inscribed angle as being defined by two chords of the circle sharing an endpoint.

Property

An inscribed angle is said to intercept an arc on the circle. The intercepted arc is the portion of the circle that is in the interior of the angle. The measure of the intercepted arc (equal to its central angle) is exactly twice the measure of the inscribed angle.

This single property has a number of consequences within the circle. For example, it allows one to prove that when two chords intersect in a circle, the products of the lengths of their pieces are equal. It also allows one to prove that the opposite angles of a cyclic quadrilateral are supplementary.

Proof

To understand this proof, it is useful to draw a diagram.

Inscribed angles where one chord is a diameter

Let "O" be the center of a circle. Choose two points on the circle, and call them "V" and "A". Draw line "VO" and extended past "O" so that it intersects the circle at point "B" which is diametrically opposite the point "V". Draw an angle whose vertex is point "V" and whose sides pass through points "A" and "B".

Angle "BOA" is a central angle; call it "θ". Draw line "OA". Lines "OV" and "OA" are both radii of the circle, so they have equal lengths. Therefore triangle "VOA" is isosceles, so angle "BVA" (the inscribed angle) and angle "VAO" are equal; let each of them be denoted as "ψ".

Angles "BOA" and "AOV" are supplementary. They add up to 180°, since line "VB" passing through "O" is a straight line. Therefore angle "AOV" measures 180° − θ.

It is known that the three angles of a triangle add up to 180°, and the three angles of triangle "VOA" are:

: 180° − θ: ψ: ψ.

Therefore

:$2\; psi\; +\; 180^circ\; -\; heta\; =\; 180^circ.$

Subtract "180°" from both sides,

:$2\; psi\; =\; heta,\; ,$

where "θ" is the central angle subtending arc "AB" and "ψ" is the inscribed angle subtending arc "AB".

Inscribed angles with the center of the circle in their interior

Given a circle whose center is point "O", choose three points "V", "C", and "D" on the circle. Draw lines "VC" and "VD": angle "DVC" is an inscribed angle. Now draw line "VO" and extend it past point "O" so that it intersects the circle at point "E". Angle "DVC" subtends arc "DC" on the circle.

Suppose this arc includes point "E" within it. Point "E" is diametrically opposite to point "V". Angles "DVE" and "EVC" are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.

Therefore

:$angle\; DVC\; =\; angle\; DVE\; +\; angle\; EVC.\; ,$

then let

:$psi\_0\; =\; angle\; DVC,$:$psi\_1\; =\; angle\; DVE,$:$psi\_2\; =\; angle\; EVC,$

so that

:$psi\_0\; =\; psi\_1\; +\; psi\_2.\; qquad\; qquad\; (1)$

Draw lines "OC" and "OD". Angle "DOC" is a central angle, but so are angles "DOE" and "EOC", and:$angle\; DOC\; =\; angle\; DOE\; +\; angle\; EOC.$

Let

:$heta\_0\; =\; angle\; DOC,$:$heta\_1\; =\; angle\; DOE,$:$heta\_2\; =\; angle\; EOC,$

so that

:$heta\_0\; =\; heta\_1\; +\; heta\_2.\; qquad\; qquad\; (2)$

From Part One we know that $heta\_1\; =\; 2\; psi\_1$ and that $heta\_2\; =\; 2\; psi\_2$. Combining these results with equation (2) yields

:$heta\_0\; =\; 2\; psi\_1\; +\; 2\; psi\_2\; ,$

therefore, by equation (1),

:$heta\_0\; =\; 2\; psi\_0.\; ,$

Inscribed angles with the center of the circle in their exterior

[The previous case can be extended to cover the case where the measure of the inscribed angle is the "difference" between two inscribed angles as discussed in the first part of this proof.]

Given a circle whose center is point "O", choose three points "V", "C", and "D" on the circle. Draw lines "VC" and "VD": angle "DVC" is an inscribed angle. Now draw line "VO" and extend it past point "O" so that it intersects the circle at point "E". Angle "DVC" subtends arc "DC" on the circle.

Suppose this arc does not include point "E" within it. Point "E" is diametrically opposite to point "V". Angles "DVE" and "EVC" are also inscribed angles, but both of these angles have one side which passes through the center of the circle, therefore the theorem from the above Part 1 can be applied to them.

Therefore:$angle\; DVC\; =\; angle\; EVC\; -\; angle\; DVE$.then let:$psi\_0\; =\; angle\; DVC,$:$psi\_1\; =\; angle\; DVE,$:$psi\_2\; =\; angle\; EVC,$so that:$psi\_0\; =\; psi\_2\; -\; psi\_1\; qquad\; qquad\; (3)$

Draw lines "OC" and "OD". Angle "DOC" is a central angle, but so are angles "DOE" and "EOC", and:$angle\; DOC\; =\; angle\; EOC\; -\; angle\; DOE.$Let:$heta\_0\; =\; angle\; DOC,$:$heta\_1\; =\; angle\; DOE,$:$heta\_2\; =\; angle\; EOC,$so that:$heta\_0\; =\; heta\_2\; -\; heta\_1\; qquad\; qquad\; (4)$

From Part One we know that $heta\_1\; =\; 2\; psi\_1$ and that $heta\_2\; =\; 2\; psi\_2$. Combining these results with equation (4) yields:$heta\_0\; =\; 2\; psi\_2\; -\; 2\; psi\_1$therefore, by equation (3),:$heta\_0\; =\; 2\; psi\_0.$

External links

* [http://www.cut-the-knot.org/pythagoras/Munching/inscribed.shtml Munching on Inscribed Angles] at cut-the-knot
* [http://www.mathopenref.com/arccentralangle.html Arc Central Angle] With interactive animation
* [http://www.mathopenref.com/arcperipheralangle.html Arc Peripheral (inscribed) Angle] With interactive animation
* [http://www.mathopenref.com/arccentralangletheorem.html Arc Central Angle Theorem] With interactive animation