如图， BEF的内角∠EBF平分线BD与外角∠AEF的平分线交于点D，过D作DH∥BC分别交EF、EB于G、H两点．下列结论：①S EBD：S FBD=BE：BF；②∠EFD=∠CFD；③HD=HF；④BH …

根据垂直定义得出∠CBD=∠CBE+∠DBE=90°，根据角平分线定义得出∠DBE=12∠FBE，求出∠CBE=12∠ABE，∠ACB=∠ECB，根据平行线的性质得出∠ABC=∠ECB，根据平行线的判定得 …

Question: Using the figure below, ∠FBD is 50∘, which is true regarding ∠EBF ? A ∠EBF=140∘ B ∠EBF=90∘ C ∠EBF=50∘ D ∠EBF=40∘

过点B作 BF∥AE ,因为 BF∥AE ,所以∠E=∠EBF.因为∠1=∠E,所以∠EBF=∠1.因为 BD⊥BE ,所以∠EBD=90°,所以∠1+∠2=180°-∠EBD=90°.而∠EBF+∠FBD=90°,又因为∠EBF=∠1,所 …

如图， BEF的内角∠EBF平分线BD与外角∠AEF的平分线交于点D，过D作DH∥BC分别交EF、EB于G、H两点．下列结论：①S EBD：S FBD =BE：BF；②∠EFD=∠CFD；③HD=HF；④BH …

2010年8月20日 · Further the EBF proteins contain an IPT/TIG domain and an atypical helix-loop-helix domain with a novel type of dimerization motif. The data presented here provide insights …

2021年3月4日 · Trên đường tròn (O;R) đặt liên tiếp các dây cung AB=BC=CD<R. AB cắt CD tại E. Tiếp tuyến tại B và D với đường tròn (O) cắt nhau tại F. Cmr: a) tam giác EBC đồng dạng …

2024年12月25日 · I tried FBD, but it's not the correct answer. Propose an efficient synthesis for the following transformation: solution, provide just one answer. Reagents in the correct order …

2008年8月1日 · EBF drives B cell differentiation by activating the Pax5 gene and other genes required for the pre-B and B cell receptors. New evidence suggests that expression of EBF in …

12.如图1，在正方形ABCD中，点F为对角线BD上一点，FE⊥AB于点E，将 EBF绕点B逆时针旋转到图2所示的位置，连接AE、DF，则在图2中，有以下说 …