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Winkeladditionssatz-Rechner

Wenden Sie das Winkeladditionspostulat an: Finden Sie einen fehlenden Winkel, wenn zwei benachbarte Winkel einen größeren Winkel bilden.

Geprüft von [email protected], Geometry Calculator Developer & Online Math Educator Zuletzt aktualisiert am April 24, 2026

Winkeladditionssatz-Rechner

Geben Sie zwei Werte ein — der fehlende dritte wird automatisch berechnet.

In Winkeladditionssatz-Rechner verwendete Formeln

∠AOC = ∠AOB + ∠BOC

In-Depth Tutorial: Winkeladditionssatz-Rechner

The Angle Addition Postulate is one of the foundational axioms of plane geometry. It says: if a point B lies in the interior of an angle ∠AOC, then the two smaller angles ∠AOB and ∠BOC together exactly fill the larger angle. As an equation:

∠AOC = ∠AOB + ∠BOC

This simple-looking statement is one of the most-used tools in geometry proofs — it lets you decompose a large angle into known smaller parts, or combine known parts to find a total. This tutorial covers what the postulate says precisely, the "between" condition, and how to use it in proofs.

The setup

Three rays share a common endpoint (vertex) O: ray OA, ray OB, and ray OC. Suppose ray OB is "between" rays OA and OC — meaning B lies in the interior of the angle formed by OA and OC.

Then ∠AOC is the "big" angle from OA to OC, and ∠AOB and ∠BOC are the two "small" angles into which OB divides it.

The postulate says: the big angle = sum of the small parts.

The "between" condition matters

The Angle Addition Postulate only applies when ray OB is BETWEEN rays OA and OC. If OB is outside the angle (on the other side of one of the rays), the postulate doesn't apply directly — you might still get sum relationships, but with different signs or arrangements.

What "between" means: drawing OB starts in the interior of ∠AOC. As you sweep from OA to OC, you pass through OB.

Three ways to use the postulate

Once the "between" condition is satisfied, the postulate gives you three computational shortcuts:

  1. Total from parts: if ∠AOB and ∠BOC are known, ∠AOC = ∠AOB + ∠BOC.
  2. One part from total + other part: ∠AOB = ∠AOC − ∠BOC.
  3. Decomposition: ∠AOC = ∠AOB + ∠BOC can be split further if more rays divide the angle.

This is the calculator's job: enter any two of the three values, get the third.

Worked example 1 — sum of parts

Ray OB lies between rays OA and OC. ∠AOB = 35°, ∠BOC = 50°. Find ∠AOC.

∠AOC = ∠AOB + ∠BOC = 35° + 50° = 85°.

Worked example 2 — find a part from total

Ray OB lies between OA and OC. ∠AOC = 120°, ∠AOB = 45°. Find ∠BOC.

∠BOC = ∠AOC − ∠AOB = 120° − 45° = 75°.

Worked example 3 — proof setup using the postulate

In a geometry proof, you might encounter:

Given: ∠AOC = 90°. ∠AOB = ∠BOC. Find ∠AOB.

Step 1: Apply Angle Addition: ∠AOC = ∠AOB + ∠BOC.
Step 2: Substitute the given equality (∠AOB = ∠BOC): 90° = 2 × ∠AOB.
Step 3: Solve: ∠AOB = 45°.

This three-step proof shows the postulate combined with the algebraic substitution property — a very common pattern in introductory geometry.

The angle bisector and Angle Addition

An angle bisector is a ray that divides an angle into two equal parts. By the Angle Addition Postulate:

If ray OB bisects ∠AOC, then ∠AOC = ∠AOB + ∠BOC = 2 × ∠AOB.

So each smaller angle is exactly half the total. This is the formal way to talk about angle bisectors — by combining the bisection property with the Angle Addition Postulate.

The Angle Subtraction Postulate

A corollary, sometimes called the "Angle Subtraction Postulate": if equal angles are subtracted from equal angles, the differences are equal. Symbolically:

If ∠AOC = ∠DEF and ∠AOB = ∠DEG (where B is between OA, OC and G is between DE, DF), then ∠BOC = ∠GEF.

This is just algebra applied to the Angle Addition Postulate — but having it named separately is useful in two-column proofs.

Multi-step decomposition

The postulate extends to more rays. If FOUR rays OA, OB, OC, OD share a vertex (with OB and OC between OA and OD), then:

∠AOD = ∠AOB + ∠BOC + ∠COD

The sum extends: any "big" angle can be decomposed into the sum of consecutive smaller angles, as long as the rays are arranged in interior order.

Where the postulate shows up in proofs

  • Triangle interior angles. The interior angle of a triangle at a vertex can be split into two sub-angles when a cevian (line from vertex to opposite side) is drawn. The full angle = sum of the parts.
  • Parallel-line angle proofs. When a transversal creates multiple sub-angles at a vertex, Angle Addition lets you piece them together.
  • Polygon decomposition. Computing interior angle sums often involves decomposing polygon vertex angles into pieces.
  • Angle bisector proofs. Showing that two halves of a bisected angle are equal uses Angle Addition + the definition of bisector.

Common mistakes

  • Ignoring the "between" condition. The postulate requires ray OB to be in the interior of ∠AOC. If OB is on the same side as OA or OC, or outside the angle entirely, the simple sum form doesn't apply.
  • Confusing with the linear pair / supplementary relation. Linear pair angles (those forming a straight line) sum to 180° — a separate concept. Angle Addition can give 180° (when ∠AOC is a straight angle) but does so only when the configuration is set up that way.
  • Adding angles that aren't at the same vertex. Angle Addition requires all three angles to share the vertex O. Angles at different vertices don't combine via this postulate.
  • Using "addition" to mean addition of degrees AND addition of ray endpoints. The postulate is about angle measures (in degrees), not about constructing rays. The "addition" is numerical.

Häufig gestellte Fragen – Winkeladditionssatz-Rechner

Wenn der Strahl OB zwischen den Strahlen OA und OC liegt, dann gilt ∠AOC = ∠AOB + ∠BOC. Geben Sie zwei der drei Winkel ein, und der fehlende wird berechnet.

Geben Sie die beiden Ihnen bekannten Werte ein. Wenn Sie den Gesamtwinkel und einen Teilwinkel kennen, geben Sie diese ein. Wenn Sie beide Teilwinkel kennen, geben Sie diese ein, um den Gesamtwinkel zu berechnen.

Es wird verwendet, um einen großen Winkel in bekannte Teile zu zerlegen oder um Winkelbeziehungen in Dreiecken, Parallelen und anderen geometrischen Figuren zu beweisen.

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