Câu hỏi của Nhan Nguyen Hong

Question

tam giác abc có i là giao điểm 3 đường phân giác. vẽ đường thẳng đi qua i cắt ab, ac và bc tại f,e,d.chứng minh bc/id+ac/ie=ab/if

bé chuột péo iu nước · Accepted Answer

Chứng minh BCID+ACIE=ABIFthe fraction with numerator cap B cap C and denominator cap I cap D end-fraction plus the fraction with numerator cap A cap C and denominator cap I cap E end-fraction equals the fraction with numerator cap A cap B and denominator cap I cap F end-fraction𝐵𝐶𝐼𝐷+𝐴𝐶𝐼𝐸=𝐴𝐵𝐼𝐹 Step 1: Áp dụng định lý sin cho tam giác AIDcap A cap I cap D𝐴𝐼𝐷và AIFcap A cap I cap F𝐴𝐼𝐹 Trong tam giác AIDcap A cap I cap D𝐴𝐼𝐷, ta có: ID=AI⋅sin(∠IAD)sin(∠IDA)cap I cap D equals the fraction with numerator cap A cap I center dot sine open paren angle cap I cap A cap D close paren and denominator sine open paren angle cap I cap D cap A close paren end-fraction𝐼𝐷=𝐴𝐼⋅sin(∠𝐼𝐴𝐷)sin(∠𝐼𝐷𝐴) Trong tam giác AIFcap A cap I cap F𝐴𝐼𝐹, ta có: IF=AI⋅sin(∠IAF)sin(∠IFA)cap I cap F equals the fraction with numerator cap A cap I center dot sine open paren angle cap I cap A cap F close paren and denominator sine open paren angle cap I cap F cap A close paren end-fraction𝐼𝐹=𝐴𝐼⋅sin(∠𝐼𝐴𝐹)sin(∠𝐼𝐹𝐴) Vì ADcap A cap D𝐴𝐷là phân giác của ∠BACangle cap B cap A cap C∠𝐵𝐴𝐶, nên ∠IAD=∠IAFangle cap I cap A cap D equals angle cap I cap A cap F∠𝐼𝐴𝐷=∠𝐼𝐴𝐹.
Do đó, ta có: IDIF=sin(∠IFA)sin(∠IDA)the fraction with numerator cap I cap D and denominator cap I cap F end-fraction equals the fraction with numerator sine open paren angle cap I cap F cap A close paren and denominator sine open paren angle cap I cap D cap A close paren end-fraction𝐼𝐷𝐼𝐹=sin(∠𝐼𝐹𝐴)sin(∠𝐼𝐷𝐴) Step 2: Áp dụng định lý sin cho tam giác BICcap B cap I cap C𝐵𝐼𝐶và AICcap A cap I cap C𝐴𝐼𝐶 Trong tam giác BICcap B cap I cap C𝐵𝐼𝐶, ta có: BCsin(∠BIC)=ICsin(∠IBC)the fraction with numerator cap B cap C and denominator sine open paren angle cap B cap I cap C close paren end-fraction equals the fraction with numerator cap I cap C and denominator sine open paren angle cap I cap B cap C close paren end-fraction𝐵𝐶sin(∠𝐵𝐼𝐶)=𝐼𝐶sin(∠𝐼𝐵𝐶) Trong tam giác AICcap A cap I cap C𝐴𝐼𝐶, ta có: ACsin(∠AIC)=ICsin(∠IAC)the fraction with numerator cap A cap C and denominator sine open paren angle cap A cap I cap C close paren end-fraction equals the fraction with numerator cap I cap C and denominator sine open paren angle cap I cap A cap C close paren end-fraction𝐴𝐶sin(∠𝐴𝐼𝐶)=𝐼𝐶sin(∠𝐼𝐴𝐶) Từ đó, ta có: BCAC=sin(∠BIC)⋅sin(∠IAC)sin(∠AIC)⋅sin(∠IBC)the fraction with numerator cap B cap C and denominator cap A cap C end-fraction equals the fraction with numerator sine open paren angle cap B cap I cap C close paren center dot sine open paren angle cap I cap A cap C close paren and denominator sine open paren angle cap A cap I cap C close paren center dot sine open paren angle cap I cap B cap C close paren end-fraction𝐵𝐶𝐴𝐶=sin(∠𝐵𝐼𝐶)⋅sin(∠𝐼𝐴𝐶)sin(∠𝐴𝐼𝐶)⋅sin(∠𝐼𝐵𝐶) Step 3: Áp dụng định lý sin cho tam giác AIEcap A cap I cap E𝐴𝐼𝐸và AIFcap A cap I cap F𝐴𝐼𝐹 Trong tam giác AIEcap A cap I cap E𝐴𝐼𝐸, ta có: IEsin(∠IAE)=AIsin(∠AEI)the fraction with numerator cap I cap E and denominator sine open paren angle cap I cap A cap E close paren end-fraction equals the fraction with numerator cap A cap I and denominator sine open paren angle cap A cap E cap I close paren end-fraction𝐼𝐸sin(∠𝐼𝐴𝐸)=𝐴𝐼sin(∠𝐴𝐸𝐼) Trong tam giác AIFcap A cap I cap F𝐴𝐼𝐹, ta có: IFsin(∠IAF)=AIsin(∠AFI)the fraction with numerator cap I cap F and denominator sine open paren angle cap I cap A cap F close paren end-fraction equals the fraction with numerator cap A cap I and denominator sine open paren angle cap A cap F cap I close paren end-fraction𝐼𝐹sin(∠𝐼𝐴𝐹)=𝐴𝐼sin(∠𝐴𝐹𝐼) Vì AIcap A cap I𝐴𝐼là phân giác của ∠Aangle cap A∠𝐴, nên ∠IAE=∠IAFangle cap I cap A cap E equals angle cap I cap A cap F∠𝐼𝐴𝐸=∠𝐼𝐴𝐹.
Do đó, ta có: IEIF=sin(∠AEI)sin(∠AFI)the fraction with numerator cap I cap E and denominator cap I cap F end-fraction equals the fraction with numerator sine open paren angle cap A cap E cap I close paren and denominator sine open paren angle cap A cap F cap I close paren end-fraction𝐼𝐸𝐼𝐹=sin(∠𝐴𝐸𝐼)sin(∠𝐴𝐹𝐼) Step 4: Sử dụng các mối quan hệ trên để chứng minh đẳng thức Từ các bước trên, ta có thể suy ra: BCID+ACIE=SABC/2SAIB/2+SABC/2SAIC/2the fraction with numerator cap B cap C and denominator cap I cap D end-fraction plus the fraction with numerator cap A cap C and denominator cap I cap E end-fraction equals the fraction with numerator cap S sub cap A cap B cap C end-sub / 2 and denominator cap S sub cap A cap I cap B end-sub / 2 end-fraction plus the fraction with numerator cap S sub cap A cap B cap C end-sub / 2 and denominator cap S sub cap A cap I cap C end-sub / 2 end-fraction𝐵𝐶𝐼𝐷+𝐴𝐶𝐼𝐸=𝑆𝐴𝐵𝐶/2𝑆𝐴𝐼𝐵/2+𝑆𝐴𝐵𝐶/2𝑆𝐴𝐼𝐶/2 =SABCSAIB+SABCSAIC=SAIC+SBIC+SAIBSAIB+SAIC+SBIC+SAIBSAICequals the fraction with numerator cap S sub cap A cap B cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap B cap C end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction equals the fraction with numerator cap S sub cap A cap I cap C end-sub plus cap S sub cap B cap I cap C end-sub plus cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap I cap C end-sub plus cap S sub cap B cap I cap C end-sub plus cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction=𝑆𝐴𝐵𝐶𝑆𝐴𝐼𝐵+𝑆𝐴𝐵𝐶𝑆𝐴𝐼𝐶=𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐵+𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐶 =1+SAICSAIB+SBICSAIB+1+SAIBSAIC+SBICSAICequals 1 plus the fraction with numerator cap S sub cap A cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus 1 plus the fraction with numerator cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denom...Câu trả lời: Chứng minh BCID+ACIE=ABIFthe fraction with numerator cap B cap C and denominator cap I cap D end-fraction plus the fraction with numerator cap A cap C and denominator cap I cap E end-fraction equals the fraction with numerator cap A cap B and denominator cap I cap F end-fraction𝐵𝐶𝐼𝐷+𝐴𝐶𝐼𝐸=𝐴𝐵𝐼𝐹 Step 1: Áp dụng định lý sin cho tam giác AIDcap A cap I cap D𝐴𝐼𝐷và AIFcap A cap I cap F𝐴𝐼𝐹 Trong tam giác AIDcap A cap I cap D𝐴𝐼𝐷, ta có: ID=AI⋅sin(∠IAD)sin(∠IDA)cap I cap D equals the fraction with numerator cap A cap I center dot sine open paren angle cap I cap A cap D close paren and denominator sine open paren angle cap I cap D cap A close paren end-fraction𝐼𝐷=𝐴𝐼⋅sin(∠𝐼𝐴𝐷)sin(∠𝐼𝐷𝐴) Trong tam giác AIFcap A cap I cap F𝐴𝐼𝐹, ta có: IF=AI⋅sin(∠IAF)sin(∠IFA)cap I cap F equals the fraction with numerator cap A cap I center dot sine open paren angle cap I cap A cap F close paren and denominator sine open paren angle cap I cap F cap A close paren end-fraction𝐼𝐹=𝐴𝐼⋅sin(∠𝐼𝐴𝐹)sin(∠𝐼𝐹𝐴) Vì ADcap A cap D𝐴𝐷là phân giác của ∠BACangle cap B cap A cap C∠𝐵𝐴𝐶, nên ∠IAD=∠IAFangle cap I cap A cap D equals angle cap I cap A cap F∠𝐼𝐴𝐷=∠𝐼𝐴𝐹.
Do đó, ta có: IDIF=sin(∠IFA)sin(∠IDA)the fraction with numerator cap I cap D and denominator cap I cap F end-fraction equals the fraction with numerator sine open paren angle cap I cap F cap A close paren and denominator sine open paren angle cap I cap D cap A close paren end-fraction𝐼𝐷𝐼𝐹=sin(∠𝐼𝐹𝐴)sin(∠𝐼𝐷𝐴) Step 2: Áp dụng định lý sin cho tam giác BICcap B cap I cap C𝐵𝐼𝐶và AICcap A cap I cap C𝐴𝐼𝐶 Trong tam giác BICcap B cap I cap C𝐵𝐼𝐶, ta có: BCsin(∠BIC)=ICsin(∠IBC)the fraction with numerator cap B cap C and denominator sine open paren angle cap B cap I cap C close paren end-fraction equals the fraction with numerator cap I cap C and denominator sine open paren angle cap I cap B cap C close paren end-fraction𝐵𝐶sin(∠𝐵𝐼𝐶)=𝐼𝐶sin(∠𝐼𝐵𝐶) Trong tam giác AICcap A cap I cap C𝐴𝐼𝐶, ta có: ACsin(∠AIC)=ICsin(∠IAC)the fraction with numerator cap A cap C and denominator sine open paren angle cap A cap I cap C close paren end-fraction equals the fraction with numerator cap I cap C and denominator sine open paren angle cap I cap A cap C close paren end-fraction𝐴𝐶sin(∠𝐴𝐼𝐶)=𝐼𝐶sin(∠𝐼𝐴𝐶) Từ đó, ta có: BCAC=sin(∠BIC)⋅sin(∠IAC)sin(∠AIC)⋅sin(∠IBC)the fraction with numerator cap B cap C and denominator cap A cap C end-fraction equals the fraction with numerator sine open paren angle cap B cap I cap C close paren center dot sine open paren angle cap I cap A cap C close paren and denominator sine open paren angle cap A cap I cap C close paren center dot sine open paren angle cap I cap B cap C close paren end-fraction𝐵𝐶𝐴𝐶=sin(∠𝐵𝐼𝐶)⋅sin(∠𝐼𝐴𝐶)sin(∠𝐴𝐼𝐶)⋅sin(∠𝐼𝐵𝐶) Step 3: Áp dụng định lý sin cho tam giác AIEcap A cap I cap E𝐴𝐼𝐸và AIFcap A cap I cap F𝐴𝐼𝐹 Trong tam giác AIEcap A cap I cap E𝐴𝐼𝐸, ta có: IEsin(∠IAE)=AIsin(∠AEI)the fraction with numerator cap I cap E and denominator sine open paren angle cap I cap A cap E close paren end-fraction equals the fraction with numerator cap A cap I and denominator sine open paren angle cap A cap E cap I close paren end-fraction𝐼𝐸sin(∠𝐼𝐴𝐸)=𝐴𝐼sin(∠𝐴𝐸𝐼) Trong tam giác AIFcap A cap I cap F𝐴𝐼𝐹, ta có: IFsin(∠IAF)=AIsin(∠AFI)the fraction with numerator cap I cap F and denominator sine open paren angle cap I cap A cap F close paren end-fraction equals the fraction with numerator cap A cap I and denominator sine open paren angle cap A cap F cap I close paren end-fraction𝐼𝐹sin(∠𝐼𝐴𝐹)=𝐴𝐼sin(∠𝐴𝐹𝐼) Vì AIcap A cap I𝐴𝐼là phân giác của ∠Aangle cap A∠𝐴, nên ∠IAE=∠IAFangle cap I cap A cap E equals angle cap I cap A cap F∠𝐼𝐴𝐸=∠𝐼𝐴𝐹.
Do đó, ta có: IEIF=sin(∠AEI)sin(∠AFI)the fraction with numerator cap I cap E and denominator cap I cap F end-fraction equals the fraction with numerator sine open paren angle cap A cap E cap I close paren and denominator sine open paren angle cap A cap F cap I close paren end-fraction𝐼𝐸𝐼𝐹=sin(∠𝐴𝐸𝐼)sin(∠𝐴𝐹𝐼) Step 4: Sử dụng các mối quan hệ trên để chứng minh đẳng thức Từ các bước trên, ta có thể suy ra: BCID+ACIE=SABC/2SAIB/2+SABC/2SAIC/2the fraction with numerator cap B cap C and denominator cap I cap D end-fraction plus the fraction with numerator cap A cap C and denominator cap I cap E end-fraction equals the fraction with numerator cap S sub cap A cap B cap C end-sub / 2 and denominator cap S sub cap A cap I cap B end-sub / 2 end-fraction plus the fraction with numerator cap S sub cap A cap B cap C end-sub / 2 and denominator cap S sub cap A cap I cap C end-sub / 2 end-fraction𝐵𝐶𝐼𝐷+𝐴𝐶𝐼𝐸=𝑆𝐴𝐵𝐶/2𝑆𝐴𝐼𝐵/2+𝑆𝐴𝐵𝐶/2𝑆𝐴𝐼𝐶/2 =SABCSAIB+SABCSAIC=SAIC+SBIC+SAIBSAIB+SAIC+SBIC+SAIBSAICequals the fraction with numerator cap S sub cap A cap B cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap B cap C end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction equals the fraction with numerator cap S sub cap A cap I cap C end-sub plus cap S sub cap B cap I cap C end-sub plus cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap A cap I cap C end-sub plus cap S sub cap B cap I cap C end-sub plus cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction=𝑆𝐴𝐵𝐶𝑆𝐴𝐼𝐵+𝑆𝐴𝐵𝐶𝑆𝐴𝐼𝐶=𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐵+𝑆𝐴𝐼𝐶+𝑆𝐵𝐼𝐶+𝑆𝐴𝐼𝐵𝑆𝐴𝐼𝐶 =1+SAICSAIB+SBICSAIB+1+SAIBSAIC+SBICSAICequals 1 plus the fraction with numerator cap S sub cap A cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denominator cap S sub cap A cap I cap B end-sub end-fraction plus 1 plus the fraction with numerator cap S sub cap A cap I cap B end-sub and denominator cap S sub cap A cap I cap C end-sub end-fraction plus the fraction with numerator cap S sub cap B cap I cap C end-sub and denom...

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