| 中文 | English | 定義 Definition | 性質 Property |
|---|---|---|---|
| 內心 I | Incentre | 角平分線交點 Intersection of angle bisectors |
內切圓圓心 Centre of incircle |
| 外心 O | Circumcentre | 垂直平分線交點 Intersection of perpendicular bisectors |
外接圓圓心 Centre of circumcircle |
| 重心 G | Centroid | 中線交點 Intersection of medians |
分中線成 2:1 Divides medians in ratio 2:1 |
| 垂心 H | Orthocentre | 高線交點 Intersection of altitudes |
位置取決於三角形類型 Position depends on triangle type |
• 三條角平分線的交點
• 是內切圓的圓心
• 到三邊距離相等(= 內切圓半徑 r)
• 公式:$\angle BIC = 90° + \dfrac{\angle A}{2}$
• Intersection of three angle bisectors
• Centre of the incircle
• Equidistant from all three sides (= inradius r)
• Formula: $\angle BIC = 90° + \dfrac{\angle A}{2}$
• 三條垂直平分線的交點
• 是外接圓的圓心
• 到三頂點距離相等(= 外接圓半徑 R)
• 位置:銳角→內部,直角→斜邊中點,鈍角→外部
• Intersection of three perpendicular bisectors
• Centre of the circumcircle
• Equidistant from all three vertices (= circumradius R)
• Position: Acute→inside, Right→midpoint of hypotenuse, Obtuse→outside
• 三條中線的交點
• 把中線分成 2:1(靠頂點段:靠中點段)
• 三角形的「重量中心」
• 坐標公式:$G = \left(\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}\right)$
• Intersection of three medians
• Divides each median in ratio 2:1 (from vertex)
• The "centre of mass" of the triangle
• Coordinate formula: $G = \left(\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}\right)$
• 三條高的交點
• 銳角三角形:垂心在內部
• 直角三角形:垂心在直角頂點
• 鈍角三角形:垂心在外部
• Intersection of three altitudes
• Acute triangle: Orthocentre is inside
• Right triangle: Orthocentre is at the right angle vertex
• Obtuse triangle: Orthocentre is outside
記憶口訣:
「內角外垂,重中垂高」
內心→角平分線
外心→垂直平分線
重心→中線
垂心→高
Memory Aid:
"Incentre = Intersection of Angle bisectors"
"Circumcentre = Circumcircle centre"
"Centroid = Centre of mass, 2:1 ratio"
"Orthocentre = Orthogonal (perpendicular) heights"
題目:△ABC 的頂點坐標為 A(2, 6)、B(8, 2)、C(4, -2),求重心坐標。
Question: Find the centroid of △ABC with vertices A(2, 6), B(8, 2), C(4, -2).
$G = \left(\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}\right)$
$= \left(\dfrac{2+8+4}{3}, \dfrac{6+2+(-2)}{3}\right) = \mathbf{\left(\dfrac{14}{3}, 2\right)}$
題目:I 是 △ABC 的內心,若 ∠BAC = 50°,求 ∠BIC。
Question: I is the incentre of △ABC. If ∠BAC = 50°, find ∠BIC.
$\angle BIC = 90° + \dfrac{\angle A}{2}$
$= 90° + \dfrac{50°}{2} = 90° + 25° = \mathbf{115°}$
題目:直角三角形的垂心在哪裡?
Question: Where is the orthocentre of a right-angled triangle?
直角三角形的垂心就在直角頂點上!
The orthocentre is at the right-angle vertex!
題目:O 是 △ABC 的外心,OA = 5 cm。求外接圓的半徑。
Question: O is the circumcentre of △ABC, OA = 5 cm. Find the circumradius.
外心到三頂點等距,即外接圓半徑!
$R = OA = OB = OC = \mathbf{5}$ cm
1. (3, 2) 2. 110° 3. 三角形外部 (Outside the triangle) 4. 2:1
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