如图,△ABC中,D,E分别是边BC,AB上的点,且∠1=∠2=∠3,若设△ABC,△EBD,△ADC的周长依次为c,c1,c2,求 20
如图,△ABC中,D,E分别是边BC,AB上的点,且∠1=∠2=∠3,若设△ABC,△EBD,△ADC的周长依次为c,c1,c2,求证c1+c2/c≤5/4追加!!!!!...
如图,△ABC中,D,E分别是边BC,AB上的点,且∠1=∠2=∠3,若设△ABC,△EBD,△ADC的周长依次为c,c1,c2,求证c1+c2/c≤5/4
追加!!!!!!!!!!!!!!!!!!!!!!!!!! 展开
追加!!!!!!!!!!!!!!!!!!!!!!!!!! 展开
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∠B=∠EDA=∠DAC,DE//AC
△ABC∽△EBD∽△ADC
△ABC,△EBD,△ADC周长为c,c1,c2即相似比为c:c1:c2
BC:BD:AC=c:c1:c2
设BC=ck,BD=c1k,AC=c2k
CD=BC-BD=(c-c1)k
CD/AC=AC/BC,(c-c1)/c2=c2/c
左边分子分母同除以c:
(1-c1/c)/(c2/c)=c2/c
设c2/c=x,c1/c=y
(1-y)/x=x,1-y=x^2,y=1-x^2
(c1+c2)/c=x+y=x+1-x^2=-(x-1/2)^2+5/4
当x=1/2,即c2/c=1/2时,
(c1+c2)/c最大为5/4
故c1+c2/c≤5/4
△ABC∽△EBD∽△ADC
△ABC,△EBD,△ADC周长为c,c1,c2即相似比为c:c1:c2
BC:BD:AC=c:c1:c2
设BC=ck,BD=c1k,AC=c2k
CD=BC-BD=(c-c1)k
CD/AC=AC/BC,(c-c1)/c2=c2/c
左边分子分母同除以c:
(1-c1/c)/(c2/c)=c2/c
设c2/c=x,c1/c=y
(1-y)/x=x,1-y=x^2,y=1-x^2
(c1+c2)/c=x+y=x+1-x^2=-(x-1/2)^2+5/4
当x=1/2,即c2/c=1/2时,
(c1+c2)/c最大为5/4
故c1+c2/c≤5/4
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