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第 19 週 · 三角學(二)立體 · 第 4 節

三垂線定理(製造直角的利器)

立體題裡最難的一步往往是「找垂直」。三垂線定理一句話搞定 — 射影垂直就推出斜線垂直,還能把「點到直線距離」化成勾股 √(高²+投影²)。配滿立體圖例。

▶ 第一關: 認清三條線(配置)
SVG 1 · the full configuration (P, O, A, L)
LPOAPOOAPA
SVG 2 · ① PO ⊥ plane (the height)
PO (foot)PO ⊥ plane= height
SVG 3 · ② OA ⊥ L (inside the plane)
(top view of the plane)LOAOA ⊥ L
▶ 第二關: 定理本體 + 求距離
SVG 4 · conclusion: PA ⊥ L (slant ⊥ L too)
LPOAPAred ⊥ at A: PA ⊥ L (theorem)
SVG 5 · PA is the point-to-line distance
LPAPA = distancePA ⊥ L → PA is the shortest distance P→L
SVG 6 · solve distance: PO=4, OA=3 → PA=5
OA = 3PO = 4PA = 5PA = √(4² + 3²) = √25 = 5
▶ 第三關: 逆定理 + 反向計算
SVG 7 · converse: PA⊥L & PO⊥plane → OA⊥L
LPOAgiven PA⊥L → green OA⊥L follows
SVG 8 · reverse: PA=5, PO=4 → OA=√(25−16)=3
OA = ?PO = 4PA = 5OA = √(5² − 4²) = √9 = 3
⚔ BOSS: 三垂線綜合應用
SVG 9 · cuboid: edge as PO, three-perpendicular inside
POL (base edge)APO⊥base