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Question Number 210235 by peter frank last updated on 03/Aug/24

Commented by Pnk2024 last updated on 03/Aug/24

$$△ADB\sim △EFB....(A-Atest)$$$$\Rightarrow \frac{y}{x}=\frac{AB}{EB}.......(C.S.S.T)$$$$again$$$$△BCA\sim △EFA$$$$\Rightarrow \frac{AE}{AB}=\frac{x}{z}........(C.S.S.T.)$$$$now\frac{AE}{AB}\times \frac{AB}{EB}=\frac{x}{z}\times \frac{y}{x}$$$$\Rightarrow \frac{AE}{EB}=\frac{y}{z}$$$$\Rightarrow \frac{AE}{EB+AE}=\frac{y}{y+z}$$$$\Rightarrow \frac{AE}{AB}=\frac{y}{y+z}.........\left(ii\right)$$$$butfromabove\frac{AE}{AB}=\frac{x}{z}$$$$\therefore \frac{y}{y+z}=\frac{x}{z}$$$$\Rightarrow yz=xy+xz$$$$dividingbyxyz$$$$\Rightarrow \frac{yz}{xyz}=\frac{xy}{xyz}+\frac{xz}{xyz}$$$$\Rightarrow \frac{1}{x}=\frac{1}{z}+\frac{1}{y}thisisproved$$$$$$

Commented by peter frank last updated on 04/Aug/24

$$\mathrm{thank}\mathrm{you}$$

Answered by mr W last updated on 04/Aug/24

Commented by mr W last updated on 04/Aug/24

$$\frac{x}{z}=\frac{a+b}{b}=1+\frac{a}{b}\Rightarrow \frac{a}{b}=\frac{x}{z}-1$$$$\frac{y}{z}=\frac{a+b}{a}=1+\frac{b}{a}\Rightarrow \frac{b}{a}=\frac{y}{z}-1$$$$\Rightarrow (\frac{x}{z}-1)(\frac{y}{z}-1)=1$$$$\Rightarrow xy=(x+y)z$$$$\Rightarrow \frac{1}{z}=\frac{1}{x}+\frac{1}{y}✓$$

Commented by peter frank last updated on 04/Aug/24

$$\mathrm{thank}\mathrm{you}$$

Answered by A5T last updated on 04/Aug/24

Commented by A5T last updated on 04/Aug/24

$$\frac{x}{y}=\frac{BE}{BA};\frac{x}{z}=\frac{EA}{BA}$$$$\Rightarrow \frac{x}{y}+\frac{x}{z}=\frac{BE+EA}{BA}=1\Rightarrow \frac{1}{y}+\frac{1}{z}=\frac{1}{x}$$

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