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Question Number 200476 by Rupesh123 last updated on 19/Nov/23

Answered by deleteduser1 last updated on 19/Nov/23

Commented by deleteduser1 last updated on 19/Nov/23

$$WLOG,letObetheoriginanda=1$$$$d=\frac{a+b}{2};e=\frac{a+c+d}{3}=\frac{a+c+\frac{a+b}{2}}{3}=\frac{3a+b+2c}{6}$$$$\mid b-1\mid =\mid c-1\mid \Rightarrow (b-1)(\stackrel{-}{b}-1)=(c-1)(\stackrel{-}{c}-1)$$$$\Rightarrow -\stackrel{-}{b}-b=-c-\stackrel{-}{c}\Rightarrow b+\frac{1}{b}=c+\frac{1}{c}\Rightarrow \frac{b^2+1}{b}=\frac{c^2+1}{c}$$$$\Rightarrow {cb}^2-{bc}^2=+b-c=bc(b-c)\Rightarrow bc=1$$$$EO\stackrel{?}{\mathrm{\perp }}CD\iff \frac{3a+b+2c}{3\stackrel{-}{a}+\stackrel{-}{b}+2\stackrel{-}{c}}=\frac{3+b+2c}{3+c+2b}=$$$$\frac{-(d-c)=-\left(\frac{a+b-2c}{2}\right)}{\stackrel{-}{d}-\stackrel{-}{c}=\frac{\stackrel{-}{a}+\stackrel{-}{b}-2\stackrel{-}{c}}{2}}=\frac{2c-1-b}{1+c-2b}$$$$\iff 0=0.Hence,EO\mathrm{\perp }CD$$

Commented by Rupesh123 last updated on 20/Nov/23

Very elegant, sir!

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