b)vẽ AtH cat BC tai F. CM AF IBCA BH BD= BE RC CCM BH. BD +t CH.CE = BC2 AABC nhọn, BD LAC; CELAB GT BDACE H. AHA BC=E KL a) AEHBcs ADHC. b) AFI BC va BH. BD = BE. BC B. C) BH BD+ CH.CE=BC² a) Xét AEHB 3 ADHC có : E=D= 90% gt) BAE =CHD (a got dôi ď) => AEHB co ADH Ĉ Cg -q) %3D 62AABC có 2da cao BD 8 CE cat nhau tai H, nen H la thille tam cua AABC. Do do AH IBC tai F hay AF I BC BDC = 90° Xét ABHF ABCD.co BEH DBC chug = -> A BHE CO ABCD (9-9) ..... > BH BC BF BD > BH. BD = BF BC ..1. C) Xet ACHF ACBE.Co CEH = CEB = 90° BCE chung -> ACHE CO.ACBE (g-g). CF O CH. CE = CE CB CE CH 11 Mã: BH. BP= BF,BC .(CMý b). > BH, BD + CH. CE = BF, BC + CE+ CB
