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Question Number 157292 by mr W last updated on 21/Oct/21

Commented by mr W last updated on 21/Oct/21

Commented by mr W last updated on 21/Oct/21

$$Q154599$$

Commented by ajfour last updated on 22/Oct/21

$$letAorigin.$$$$c=d+b$$$$y=\lambda dx=\mu b$$$$eq.ofXY$$$$r=\frac{\lambda d+\mu b}{2}+\rho (\mu b-\lambda d)$$$$letr=z$$$$\Rightarrow nowc=\xi z$$$$d+b=\xi \{\frac{\lambda d+\mu b}{2}+\rho (\mu b-\lambda d)\}$$$$\Rightarrow (\xi -2\rho )\lambda =2\mathrm{\&}$$$$(\xi +2\rho )\mu =2$$$$\Rightarrow \frac{2}{\lambda }-\xi +\frac{2}{\mu }-\xi =0$$$$\Rightarrow \frac{1}{\lambda }+\frac{1}{\mu }=\xi$$$$\frac{d}{y}+\frac{b}{x}=\frac{c}{z}orproperly$$$$\frac{AD}{AY}+\frac{AB}{AX}=\frac{AC}{AZ}★$$$$$$

Commented by ajfour last updated on 22/Oct/21

https://youtu.be/BeqXQ4bPv14

Commented by ajfour last updated on 22/Oct/21

$$justmyanotherdancecover!$$

Commented by mr W last updated on 22/Oct/21

$$greatsolutionandgreatdance!$$

Answered by mr W last updated on 21/Oct/21

Commented by mr W last updated on 22/Oct/21

$$AD=BC$$$$\mathrm{\Delta }ADQ\equiv \mathrm{\Delta }CBP$$$$\Rightarrow AQ=CP$$$$\frac{AB}{AX}=\frac{AP}{AZ}$$$$\frac{AD}{AY}=\frac{AQ}{AZ}=\frac{PC}{AZ}$$$$\frac{AB}{AX}+\frac{AD}{AY}=\frac{AP}{AZ}+\frac{PC}{AZ}=\frac{AP+PC}{AZ}=\frac{AC}{AZ}✓$$

Commented by Tawa11 last updated on 21/Oct/21

$$\mathrm{Great}\mathrm{sir}$$

Commented by ajfour last updated on 22/Oct/21

$$really$$

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