Question-219918

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Question Number 219918 by Spillover last updated on 03/May/25

Commented by Spillover last updated on 04/May/25

Commented by Spillover last updated on 04/May/25

Answered by mr W last updated on 04/May/25

Commented by mr W last updated on 04/May/25

say AC=b, BC=a, AB=c CL=m=((√(2a^2 +2b^2 −c^2 ))/2) CH=HG=GL=(m/3) cos α=((b^2 +m^2 −((c/2))^2 )/(2bm))=((4b^2 +4m^2 −c^2 )/(8bm)) EC=((CH)/(cos α))=(m/3)×((8bm)/(4b^2 +4m^2 −c^2 )) ((AC)/(EC))=((3(4b^2 +4m^2 −c^2 ))/(8m^2 )) similarly ((BC)/(FC))=((3(4a^2 +4m^2 −c^2 ))/(8m^2 )) ⇒((AC)/(EC))+((BC)/(FC))=((3(4a^2 +4b^2 +8m^2 −2c^2 ))/(8m^2 )) =((3(2a^2 +2b^2 +4m^2 −c^2 ))/(4m^2 )) =((3(2a^2 +2b^2 +2a^2 +2b^2 −c^2 −c^2 ))/(2a^2 +2b^2 −c^2 )) =3×2=6 ✓

$${say}\:{AC}={b},\:{BC}={a},\:{AB}={c} \\ $$$${CL}={m}=\frac{\sqrt{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${CH}={HG}={GL}=\frac{{m}}{\mathrm{3}} \\ $$$$\mathrm{cos}\:\alpha=\frac{{b}^{\mathrm{2}} +{m}^{\mathrm{2}} −\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} }{\mathrm{2}{bm}}=\frac{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{8}{bm}} \\ $$$${EC}=\frac{{CH}}{\mathrm{cos}\:\alpha}=\frac{{m}}{\mathrm{3}}×\frac{\mathrm{8}{bm}}{\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$$\frac{{AC}}{{EC}}=\frac{\mathrm{3}\left(\mathrm{4}{b}^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{\mathrm{8}{m}^{\mathrm{2}} } \\ $$$${similarly} \\ $$$$\frac{{BC}}{{FC}}=\frac{\mathrm{3}\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{\mathrm{8}{m}^{\mathrm{2}} } \\ $$$$\Rightarrow\frac{{AC}}{{EC}}+\frac{{BC}}{{FC}}=\frac{\mathrm{3}\left(\mathrm{4}{a}^{\mathrm{2}} +\mathrm{4}{b}^{\mathrm{2}} +\mathrm{8}{m}^{\mathrm{2}} −\mathrm{2}{c}^{\mathrm{2}} \right)}{\mathrm{8}{m}^{\mathrm{2}} } \\ $$$$\:\:\:=\frac{\mathrm{3}\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{4}{m}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{\mathrm{4}{m}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\frac{\mathrm{3}\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} −{c}^{\mathrm{2}} \right)}{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\mathrm{3}×\mathrm{2}=\mathrm{6}\:\checkmark \\ $$

Commented by Spillover last updated on 04/May/25

$${thank}\:{you}.{correct} \\ $$

Answered by Spillover last updated on 04/May/25

Answered by vnm last updated on 04/May/25

let CM=1 be the median of triangle ABC. ∡ACM=α, ∡BCM=β CE=(1/(3cosα)), CF=(1/(3cosβ)) CA=((2sinβ)/(sin(α+β))), CB=((2sinα)/(sin(α+β))) ((CA)/(CE))+((CB)/(CF))=((6sinβcosα)/(sin(α+β)))+((6sinαcosβ)/(sin(α+β)))=6

$$\mathrm{let}\:{CM}=\mathrm{1}\:\mathrm{be}\:\mathrm{the}\:\mathrm{median}\:\mathrm{of}\:\mathrm{triangle}\:{ABC}. \\ $$$$\measuredangle{ACM}=\alpha,\:\measuredangle{BCM}=\beta \\ $$$${CE}=\frac{\mathrm{1}}{\mathrm{3cos}\alpha},\:\:{CF}=\frac{\mathrm{1}}{\mathrm{3cos}\beta} \\ $$$${CA}=\frac{\mathrm{2sin}\beta}{\mathrm{sin}\left(\alpha+\beta\right)},\:\:{CB}=\frac{\mathrm{2sin}\alpha}{\mathrm{sin}\left(\alpha+\beta\right)} \\ $$$$\frac{{CA}}{{CE}}+\frac{{CB}}{{CF}}=\frac{\mathrm{6sin}\beta\mathrm{cos}\alpha}{\mathrm{sin}\left(\alpha+\beta\right)}+\frac{\mathrm{6sin}\alpha\mathrm{cos}\beta}{\mathrm{sin}\left(\alpha+\beta\right)}=\mathrm{6} \\ $$

Commented by Spillover last updated on 04/May/25