角度加法公理计算器
结果
角度加法公理计算器 中使用的公式
In-Depth Tutorial: 角度加法公理计算器
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:
- Total from parts: if ∠AOB and ∠BOC are known, ∠AOC = ∠AOB + ∠BOC.
- One part from total + other part: ∠AOB = ∠AOC − ∠BOC.
- 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.
常见问题解答 – 角度加法公理计算器
如果射线 OB 位于射线 OA 和 OC 之间,则 ∠AOC = ∠AOB + ∠BOC。输入三个角中的任意两个,缺失的一个将被计算。
输入您已知的两个。如果您知道总角和一个部分,输入这些。如果您知道两个部分,输入它们以求总角。
它用于将大角分解为已知部分,或证明三角形、平行线和其他几何图形中的角度关系。
是的——免费且无限制。