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Question Number 176821 by Ar Brandon last updated on 27/Sep/22

Commented by som(math1967) last updated on 27/Sep/22

 ((AF)/(AL))=((((BC)/2)+CE+((EK)/2))/(((BC)/2)+CE+EK+((KM)/2)))    =((.5+2+2)/(.5+2+4+8))=((4.5)/(14.5))=(9/(29))

$$\:\frac{{AF}}{{AL}}=\frac{\frac{{BC}}{\mathrm{2}}+{CE}+\frac{{EK}}{\mathrm{2}}}{\frac{{BC}}{\mathrm{2}}+{CE}+{EK}+\frac{{KM}}{\mathrm{2}}} \\ $$$$\:\:=\frac{.\mathrm{5}+\mathrm{2}+\mathrm{2}}{.\mathrm{5}+\mathrm{2}+\mathrm{4}+\mathrm{8}}=\frac{\mathrm{4}.\mathrm{5}}{\mathrm{14}.\mathrm{5}}=\frac{\mathrm{9}}{\mathrm{29}} \\ $$

Commented by Ar Brandon last updated on 27/Sep/22

Thank you Sir

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