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Question Number 210533 by hardmath last updated on 11/Aug/24

Commented by mr W last updated on 12/Aug/24

$$typoAD\cdot \frac{\cancel{m}n}{m+n}-AB\cdot \frac{q}{p+q}?$$

Answered by mr W last updated on 13/Aug/24

$$\frac{p}{q}\times \frac{CL}{AL}\times \frac{AJ}{BJ}=1$$$$\Rightarrow \frac{p}{q}\times \frac{AC-AL}{AL}\times \frac{AJ}{AB+AJ}=1$$$$\Rightarrow \frac{p}{p+q}\times \frac{AC}{AL}-\frac{AB}{AJ}\times \frac{q}{p+q}=1$$$$let\frac{AL}{AJ}=k$$$$\Rightarrow \frac{p}{p+q}\times AC-k\times AB\times \frac{q}{p+q}=AL...\left(i\right)$$$$$$$$\frac{m}{n}\times \frac{EL}{AL}\times \frac{AJ}{DJ}=1$$$$\Rightarrow \frac{m}{n}\times \frac{AE-AL}{AL}\times \frac{AJ}{AD+AJ}=1$$$$\Rightarrow \frac{m}{m+n}\times \frac{AE}{AL}-\frac{AD}{AJ}\times \frac{n}{m+n}=1$$$$\Rightarrow \frac{m}{m+n}\times AE-k\times AD\times \frac{n}{m+n}=AL...\left(ii\right)$$$$$$$$from\left(i\right)and\left(ii\right):$$$$\frac{p}{p+q}\times AC-k\times AB\times \frac{q}{p+q}=\frac{m}{m+n}\times AE-k\times AD\times \frac{n}{m+n}$$$$\Rightarrow k=\frac{AE\times \frac{m}{m+n}-AC\times \frac{p}{p+q}}{AD\times \frac{n}{m+n}-AB\times \frac{q}{p+q}}=\frac{AL}{AJ}✓$$

Commented by hardmath last updated on 12/Aug/24

$$\mathrm{Perfect}\mathrm{dear}\mathrm{professor}\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much}$$

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