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在三角形ABC中,三条内角平分线AD,BE,CF相交于点G,G...

证明:∵∠AEG=∠EBC+∠ACB= 1/2∠ABC+∠ACB,∴∠AGE=180°-(∠DAC+∠AEG)=180°-[1/2∠BAC+1/2∠ABC+∠ACB]=180°-[1/2(∠BAC+∠ABC)+∠ACB]=180°-[1/2 (180°-∠ACB)+∠ACB]=180°-[90°+1/2∠ACB]=90°-1/2∠ACB,∴∠BGD=∠AGH=90°-1/2∠ACB,又∵在直角△GCH中,∠CGH=90°-∠GCD=90°-1/2 ∠ACB,∴∠BGD=∠CGH.希望采纳

证明:∵∠AEG=∠EBC+∠ACB= 1/2∠ABC+∠ACB,∴∠AGE=180°-(∠DAC+∠AEG)=180°-[1/2∠BAC+1/2∠ABC+∠ACB]=180°-[1/2(∠BAC+∠ABC)+∠ACB]=180°-[1/2 (180°-∠ACB)+∠ACB]=180°-[90°+1/2∠ACB]=90°-1/2∠ACB,∴∠BGD=∠AGH=90°-1/2∠ACB,又∵在直角△GCH中,∠CGH=90°-∠GCD=90°-1/2 ∠ACB,∴∠BGD=∠CGH.好久没写了,希望采纳

解:∵∠ABC=32, ∠ACB=76∴∠BAC=180-∠ABC-∠ACB=72∵AD平分∠BAC∴∠CAD=∠BAC/2=36∵CF平分∠ACB∴∠ACF=∠BCF=∠ACB/2=38∴∠COD=∠CAD+∠ACF=36+38=74∵OG⊥BC∴∠COG+∠BCF=90∴∠COG=90-∠BCF=90-38=52∴∠DOG=∠COD-∠COG=74-52=22°

∠BOD=∠BAO+∠ABO=1/2(∠A+∠B)∠COG=90°-∠OCG=1/2(180°-∠C)=1/2(∠A+∠B)∴∠BOD=∠COG

∠DOH=(180°-∠OHB)-∠OBH-∠BOD=90°-∠OBH-∠BOD=90°-∠OBH-(∠BAD+∠ABO)=90°-1/2∠ABC-(1/2∠BAC+1/2∠ABC)=90°-∠ABC-1/2∠BAC=(90°-1/2∠BAC)-∠ABC=1/2(180°-∠BAC)-∠ABC=1/2(∠ABC+∠ACB)-∠ABC=1/2(∠ACB-∠ABC)

角BAD和角DAC,因为角EGC是三角形BCG的外角,所以角EGC等于角ECB加角BCG,又因为三角形的内角和是180度,,且因为三条角平分线,所以角BCG加角GBC加角BAD或DAC相加等于90度,又因为角EGC等于角GBC加角GCB,所以角EGC加角BAD或角DAC等于90度,也就是互余,在iPad上打这么多字不容易,给个最佳吧

是角CAD 与角BAD根据三角形的内角和等于180 ,题中角平分线的条件,可知∠EBC+∠GCB+∠CAD=90 (这三个角都等于大角的一半)∠EBC+∠GCB=∠EGC 所以∠EGC+∠CAD=90 此时∠EGC与∠CAD互余 ∠CAD=∠BAD 所以∠EGC与∠CAD ∠BAD 均互余

等于由题可知:∠BAD+∠CAD+∠EAG+∠ABG=90° 因为2(∠ABG+∠BAD+∠ACF)=180° 所以∠BAD+∠CAD+∠EAG+∠ABG=∠BAD+∠ACF+∠ABG 即∠CAD+∠EAG=∠ACF 因此∠HAG=1/ 2∠ACB

没图也没关系的啦 O(∩_∩)O~由对顶角AHF=CHD=180-(HCD CDH)(1)又因为CDH=DAB DBA=DAB 2HBG(2)将(2)带入(1)得AHF=180-(HCD DAB 2HBG)(3)又因为三角形内角和是180所以各个内角的一半加起来是90所以HCD DAB HBG=90带入(3)得AHF=180-(90 HBG)=90-HBG=BHG

∠BGD=∠GAB+∠GBA =∠BAC的一半+∠ABC的一半 =(∠BAC+∠ABC)/2 =(180-∠ACB)/2 =90-∠ACB的一半 =90-∠GCB =∠CGH

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