最重要就係要注意到 將圖6(b)攤返平之後 FH會同AC垂直
注意到呢一點之後剩低嘅就係2D問題
以下都係講緊將圖6(b)攤返平之後嘅情況
AC=sqrt(AB^2 + AC^2)=36.05551275
設P為由G至AC的垂足。
留意ΔGPC與ΔABC相似。
GP/AB = GC/AC = PC/BC
=> GP/20 = (19.29632483/2)/36.05551275 = PC/30
=> GP = 5.351837585 及 PC = 8.027756377
AP = AC - PC = 28.02775638
AG = sqrt(AB^2 + BG^2) = sqrt(AB^2 + (BE+EG)^2) = 28.53414259
設Q為FH與AC之間嘅交點
AQ = AF*sin(∠FAQ) = 16.05551275
留意ΔAHQ與ΔAGP相似。
AH/AG = AQ/AP
=> AH/28.53414259 = 16.05551275/28.02775638
=> AH = 16.34559271