初三几何题（相似形相关）

∵DE∥AB,DF∥AC
∴△FBD∽△EDC∽△ABC
S△FBD=BD²/BC²*S△ABC
S△EDC=DC²/BC²*S△ABC
S△FBD+S△EDC+S□AFDE=S△ABC
BD²/BC²*S△ABC+DC²/BC²*S△ABC+12/25S△ABC=S△ABC
BD²/BC²+DC²/BC²=13/25
25(BD²+DC²)=13BC²=13(BD+DC)²
6BD-13BD*DC+6DC²=0
(3BD-2DC)(2BD-3DC)=0
∴BD/DC=3/2,或BD/DC=3/2