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-GF=HG，其中正确结论的个数有（ ） A．只有①②③ B．只有①②④ C．只有③④ D．①②③④ A EH GD BF C

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出AC∥BE，根据三角形的内角和定理得出∠BCD+∠D=90°，即可得出答案．. ①∵BC⊥BD，∴∠DBE+∠CBE=90∘,∠ABC+∠DBF=90∘，又∵BD平分∠EBF，∴∠DBE=∠DBF，∴∠ABC=∠CBE，即BC平分∠ABE，正确；②由AB∥CE,BC平 …

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,所以∠FBD=∠2.因为∠2=∠D,所以∠FBD=∠D,所以 BF∥CD 所以 AE∥CD

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-GF=HG，其中正确结论的个数有（ ）

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 into unique structural features of the EBF proteins and open possibilities for detailed molecular investigations of this important transcription factor family.

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 tam giác FBD b) tam giác EBF đồng dạng tam giác CBD C, chứng minh BC // EF

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 as a string of letters without spaces or punctuation, such as EBF. If there is more than one correct: A. TsCl, pyridine B. 1CO2HO, CHCOCI, pyridine C. 1O=CH2HO E. CHCONa ...

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 common lymphoid progenitors directs B cell fate decisions.

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，有以下说法：①FD=AE；②∠AEB=135°；③=1：2；④AE∥BF，其中正确结论的序号是（ ）A DA DE FF EB CB图2C图1 A. ①② B.①③ C.②③ D.③④