Theorem of the inscribed angle and the angle in the center

Theorem of the inscribed angle and the angle in the center
In a Circle, an angle in the center measures the double of an inscribed angle which intercepts same the arc.

The theorem of the angle and the angle in the center is a Théorème of Euclidean Géométrie which exists two forms, one concerning the geometrical angles and the other concerning the angles orientés.

Theorem of the inscribed angle
Two inscribed angles in a circle intercepting the same arc have same the mesure.

There exist two theorems concerning this property, one relating to the geometrical angles and the other relating one to the directed angles.

Theorem of the inscribed angle and the angle in the center

If M is a point of a circle Γ, center O, if has and B are two points of the circle then: $2 \ widehat {AMB} = \ widehat {AOB}$ .

There thus exist two situations, one where the inscribed angle of top M is acute, therefore l" angle in the center of top O projecting (figure 1), the other where the inscribed angle of top M is blunt, therefore the angle in the center of top O returning (figure 2).

Demonstration with the geometrical angles

The demonstration of this theorem is done in two stages.

Initially (figure above on the left) one shows that if is a diameter then one a: $2 \ widehat {AMD} = \ widehat {AOD}$ .

Indeed, one a:

\ widehat {AOD} = 180^ \ circ - \ widehat {AOM}

and as the triangle AOM is isosceles of top O, one knows that:

180^ \ circ - \ widehat {AOM} = 2 \ widehat {AMD}

From where equality.

In other time one notices that, whatever the positions of has and B the angle $\ widehat {AMB}$ is the sum (figure of the center) or the difference (figure of right-hand side) of the angles $\ widehat {AMD}$ and $\ widehat {DMB}$ and that it will be the same for the angle $\ widehat {AOB}$ , nap or the difference of the angles $\ widehat {AOD}$ and $\ widehat {DOB}$ .

Demonstration with the directed angles

The statement and the demonstration of the property are much simpler with angles orientés.
If has and B are two points of a circle Γ of center O and if M is a point of Γ distinct from has and B then:

(\ overrightarrow {OA}, \ overrightarrow {OB}) \ equiv 2 (\ overrightarrow {MY}, \ overrightarrow {MB}) \ MOD {2 \ pi}

the demonstration uses simply the Relation of Chasles on the directed angles and the property of the isosceles triangles.

(\ overrightarrow {OA}, \ overrightarrow {OB}) \ equiv (\ overrightarrow {OA}, \ overrightarrow {OM}) + (\ overrightarrow {OM}, \ overrightarrow {OB}) \ MOD {2 \ pi}

As triangles OAM and OBM are isosceles, one a: $(\ overrightarrow {OA}, \ overrightarrow {OM}) \ equiv \ pi - 2 (\ overrightarrow {MO}, \ overrightarrow {MY}) \ MOD {2 \ pi}$ and: $(\ overrightarrow {OM}, \ overrightarrow {OB}) \ equiv \ pi - 2 (\ overrightarrow {MB}, \ overrightarrow {MO}) \ MOD {2 \ pi}$ .

While replacing one obtains: $(\ overrightarrow {OA}, \ overrightarrow {OB}) \ equiv 2 \ pi - 2 ((\ overrightarrow {MB}, \ overrightarrow {MO}) + (\ overrightarrow {MO}, \ overrightarrow {MY})) \ MOD {2 \ pi}$

(\ overrightarrow {OA}, \ overrightarrow {OB}) \ equiv - 2 (\ overrightarrow {MB}, \ overrightarrow {MY}) \ MOD {2 \ pi}

(\ overrightarrow {OA}, \ overrightarrow {OB}) \ equiv 2 (\ overrightarrow {MY}, \ overrightarrow {MB}) \ MOD {2 \ pi}

reciprocal Property :
If has and B are two points distinct from a circle Γ of center O and if: $(\ overrightarrow {OA}, \ overrightarrow {OB}) \ equiv 2 (\ overrightarrow {MY}, \ overrightarrow {MB}) \ MOD {2 \ pi}$ then M is on the circle.

This property is shown by noticing that the preceding equality prevent the points M , has and B to be aligned (the angle $(\ overrightarrow {OA}, \ overrightarrow {OB})$ is never null). One can thus consider the O' center of the circle circumscribed with the triangle MAB and use the direct direction of the property

(\ overrightarrow {O' has}, \ overrightarrow {O' B}) \ equiv 2 (\ overrightarrow {MY}, \ overrightarrow {MB}) \ MOD {2 \ pi}

one thus obtains: $(\ overrightarrow {O' has}, \ overrightarrow {O' B}) \ equiv (\ overrightarrow {OA}, \ overrightarrow {OB}) \ MOD {2 \ pi}$ .

The isosceles triangles (OAB) and (O' AB) even have bases and even point angle, they are thus confused and O' = O . The point M is of course circle Γ.

Theorem of the inscribed angle

Demonstration for geometrical angles

Two inscribed angles in a Cercle intercepting the same arc of circle have same measurement.

A Angle is registered in a Cercle if its Sommet belongs to the circle. The arc which it intercepts can be outgoing or returning. In the second case, the geometrical angles are blunt but the property is stated in the same way: $\ widehat {AMB} = \ widehat {ANB}$ .

This property is a direct consequence of the Théorème of the inscribed angle and angle in the center.
Indeed, since: $\ widehat {AMB} = \ frac 12 \ widehat {AOB}$ and $\ widehat {ANB} = \ frac 12 \ widehat {AOB}$ it comes immediately that: $\ widehat {AMB} = \ widehat {ANB}$ .

Demonstration for directed angles

For the directed angles, the property becomes a characterization of the circle passing by the points AMB .

If three points has , M , B is not aligned, and if $(\ Gamma)$ is the circle circumscribed at the three points then for any point NR distinct from has and B, there is $(\ overrightarrow {NA}, \ overrightarrow {NB}) \ equiv (\ overrightarrow {MY}, \ overrightarrow {MB}) \ MOD \ pi \ yew NR \ in (\ Gamma)$ .

It will be noticed that the equality is not true that with π close what explains why the geometrical angles can be different.

Angle of the cord and a tangent

The property of the inscribed angles spreads with the angles which the cord forms which underlies the arc with a tangent:
The inscribed angle equal to the one of the two angles is formed by a cord, which joint ends of the arc, with one of the tangents to the circle with the one of these extrémités.
.
The inscribed angle

\ widehat {AMB}

equal to one two angles is formed by tangent (TT') with the circle in has with the cord:

The inscribed angle $\ widehat {AMB}$ has even measurement that the angle $\ widehat {BEATS}$ cord with the tangent

The inscribed angle $\ widehat {ANB}$ has even measurement that the angle $\ widehat {BAT'}$ of the cord with the tangent

$\ widehat {BEATS}$ is the limiting position of the inscribed angle $\ widehat {BMA}$ when M " tend" towards A.

Demonstration :
If H is the medium of, the angles $\ widehat {HOA}$ and $\ widehat {BEATS}$ have their sides two to two perpendiculars, they are égaux.
(OH) being the bisectrix of the isosceles triangle BOA, one has $\ widehat {HOA} = \ frac 1 2 \ widehat {BOA}$ and $\ widehat {BEATS}$ is quite equal to half of the angle in the center $\ widehat {BOA}$ .

See too

Category: Ring and sphere Category: Angle

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