Question Number 210533 by hardmath last updated on 11/Aug/24
Commented by mr W last updated on 12/Aug/24
$${typo}\:{AD}\centerdot\frac{\cancel{{m}}\:{n}}{{m}+{n}}−{AB}\centerdot\frac{{q}}{{p}+{q}}\:? \\ $$
Answered by mr W last updated on 13/Aug/24
$$\frac{{p}}{{q}}×\frac{{CL}}{{AL}}×\frac{{AJ}}{{BJ}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{p}}{{q}}×\frac{{AC}−{AL}}{{AL}}×\frac{{AJ}}{{AB}+{AJ}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{p}}{{p}+{q}}×\frac{{AC}}{{AL}}−\frac{{AB}}{{AJ}}×\frac{{q}}{{p}+{q}}=\mathrm{1} \\ $$$${let}\:\frac{{AL}}{{AJ}}={k} \\ $$$$\Rightarrow\frac{{p}}{{p}+{q}}×{AC}−{k}×{AB}×\frac{{q}}{{p}+{q}}={AL}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$$\frac{{m}}{{n}}×\frac{{EL}}{{AL}}×\frac{{AJ}}{{DJ}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{m}}{{n}}×\frac{{AE}−{AL}}{{AL}}×\frac{{AJ}}{{AD}+{AJ}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{m}}{{m}+{n}}×\frac{{AE}}{{AL}}−\frac{{AD}}{{AJ}}×\frac{{n}}{{m}+{n}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{m}}{{m}+{n}}×{AE}−{k}×{AD}×\frac{{n}}{{m}+{n}}={AL}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$$\frac{{p}}{{p}+{q}}×{AC}−{k}×{AB}×\frac{{q}}{{p}+{q}}=\frac{{m}}{{m}+{n}}×{AE}−{k}×{AD}×\frac{{n}}{{m}+{n}} \\ $$$$\Rightarrow{k}=\frac{{AE}×\frac{{m}}{{m}+{n}}−{AC}×\frac{{p}}{{p}+{q}}}{{AD}×\frac{{n}}{{m}+{n}}−{AB}×\frac{{q}}{{p}+{q}}}=\frac{{AL}}{{AJ}}\:\checkmark \\ $$
Commented by hardmath last updated on 12/Aug/24
$$\mathrm{Perfect}\:\mathrm{dear}\:\mathrm{professor}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$