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Question Number 23592 by Tinkutara last updated on 02/Nov/17

$$\mathrm{Let}ABC\mathrm{be}\mathrm{a}\mathrm{triangle}\mathrm{with}AB=AC$$$$\mathrm{and}\mathrm{\angle }BAC=30°.\mathrm{Let}{A}^{\prime }\mathrm{be}\mathrm{the}\mathrm{reflection}$$$$\mathrm{of}A\mathrm{in}\mathrm{the}\mathrm{line}BC;B^{\prime }\mathrm{be}\mathrm{the}\mathrm{reflection}$$$$\mathrm{of}B\mathrm{in}\mathrm{the}\mathrm{line}CA;C^{\prime }\mathrm{be}\mathrm{the}\mathrm{reflection}$$$$\mathrm{of}C\mathrm{in}\mathrm{the}\mathrm{line}AB.\mathrm{Show}\mathrm{that}{A}^{\prime },B^{\prime },C^{\prime }$$$$\mathrm{form}\mathrm{the}\mathrm{vertices}\mathrm{of}\mathrm{an}\mathrm{equilateral}$$$$\mathrm{triangle}.$$

Answered by ajfour last updated on 02/Nov/17

Commented by Tinkutara last updated on 06/Nov/17

Commented by Tinkutara last updated on 06/Nov/17

Commented by ajfour last updated on 02/Nov/17

$$Given\mathit{\theta }=15°,toprove\mathit{\varphi }=30°$$

Commented by ajfour last updated on 02/Nov/17

$$OB{}^{\prime }=BB{}^{\prime }-OB$$$$=2a\cos \theta -\frac{a}{2}\sec \theta$$$$MB{}^{\prime }=OB{}^{\prime }\cos \theta$$$$=2a\cos {}^2\theta -\frac{a}{2}=\frac{a}{2}(4\cos {}^2\theta -1)$$$$A^{\prime }M=OM+OD+A{}^{\prime }D$$$$A^{\prime }D=AD=\frac{a}{2}\cot \theta ,so$$$$A^{\prime }M=OB{}^{\prime }\sin \theta +\frac{a}{2}\tan \theta +\frac{a}{2}\cot \theta$$$$=(2a\cos \theta -\frac{a}{2}\sec \theta )\sin \theta$$$$+\frac{a}{2}\tan \theta +\frac{a}{2}\cot \theta$$$$=\frac{a}{2}(4\sin \theta \cos \theta +\cot \theta )$$$$\tan \varphi =\frac{MB{}^{\prime }}{A^{\prime }M}$$$$=\frac{\frac{a}{2}(4\cos {}^2\theta -1)}{\frac{a}{2}(4\sin \theta \cos \theta +\cot \theta )}$$$$=\frac{\frac{a}{2}(2+2\cos 2\theta -1)}{\frac{a}{2}(2\sin 2\theta +\sqrt{\frac{1+\cos 2\theta }{1-\cos 2\theta }})}$$$$given\theta =15°\Rightarrow 2\theta =30°,so$$$$\tan \varphi =\frac{2+\sqrt{3}-1}{1+\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}}=\frac{\sqrt{3}+1}{1+2+\sqrt{3}}$$$$=\frac{1}{\sqrt{3}}.$$$$\Rightarrow \mathit{\varphi }=30°andbysymmetry$$$$A^{\prime }C{}^{\prime }=A^{\prime }B{}^{\prime }$$$$Hence△A^{\prime }B{}^{\prime }C{}^{\prime }is\mathit{e}\mathit{q}\mathit{u}\mathit{i}\mathit{l}\mathit{a}\mathit{t}\mathit{e}\mathit{r}\mathit{a}\mathit{l}.$$$$$$

Commented by Tinkutara last updated on 02/Nov/17

$$Why{AA}^{\prime }bisects\mathrm{\angle }{C}^{\prime }{A}^{\prime }{B}^{\prime }?$$

Commented by ajfour last updated on 02/Nov/17

$$honestly,idontknow.$$

Commented by Tinkutara last updated on 06/Nov/17

Commented by ajfour last updated on 06/Nov/17

$$usingwhichapp.haveyoutyped$$$$anddrawnallthese?$$

Commented by Tinkutara last updated on 15/Nov/17

$$Sir,canyoushowthatBC\parallel B^{\prime }{C}^{\prime }?$$$$BecauseIwanttoknowhowyou$$$$assumed\mathrm{\angle }{AMC}^{\prime }=90°.$$

Commented by ajfour last updated on 15/Nov/17

$$letxaxisbealongBCwithD$$$$asorigin.Since△ABCis$$$$isosceles,\mathrm{\angle }OBC=\mathrm{\angle }OCB$$$$furtherOB=OC$$$$andBE=CF$$$$so2BE=2CF$$$$\Rightarrow BB{}^{\prime }=CC{}^{\prime }=l\left(say\right)$$$$Hence{y}_{B{}^{\prime }}=y_{C{}^{\prime }}=l\sin \theta$$$$henceB{}^{\prime }C{}^{\prime }//BC.$$$$AndAD\mathrm{\perp }BC$$$$\Rightarrow \mathrm{\angle }AMC{}^{\prime }=\mathrm{\angle }AMB{}^{\prime }=90°.$$

Commented by Tinkutara last updated on 15/Nov/17

$$Whatis{y}_{B^{\prime }}?$$

Commented by ajfour last updated on 15/Nov/17

$$Isaid,letstakeBCasxaxis$$$$withDasorigin.$$$$y_{B{}^{\prime }}isycoordinateofB{}^{\prime }then.$$

Commented by Tinkutara last updated on 15/Nov/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

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