如圖AC=CD=30,O為直角三角形ACD的外心,G為直角三角形CED的重心求OG線段的長?

若O為直角三角形ACB的外心:
△ACB ≌ △CDE (ASA)
設o在BC上使oO⊥BC, 則 oC = BC/2 = √(30² - 18²)/2 = 12, oO = AB/2 = 9.
設g在CE上使gG⊥CE, 則 gC = (1+1/3)(CE/2) = (4/3)(18/2) = 12, gG = (1-1/3)(ED/2) = (2/3)(24/2) = 8.
OG² = (oO - gG)² + (oC + gC)²
OG² = (9 - 8)² + (12 + 12)²
OG = √577