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Question Number 211672 by MATHEMATICSAM last updated on 15/Sep/24

$$\mathrm{In}\mathrm{a}\mathrm{triangle}\mathrm{the}\mathrm{bisector}\mathrm{of}\mathrm{the}\mathrm{side}c\mathrm{is}$$$$\mathrm{perpendicular}\mathrm{to}\mathrm{side}b.\mathrm{Prove}\mathrm{that}$$$$2\mathrm{tanC}+\mathrm{tanA}=0.$$

Commented by Frix last updated on 16/Sep/24

$$\mathrm{If}\mathrm{the}\mathrm{bisector}\mathrm{of}\mathrm{a}side\mathrm{is}\mathrm{perpendicular}\mathrm{to}$$$$\mathrm{the}\mathrm{other}\mathrm{side}\Rightarrow \mathrm{the}\mathrm{angle}\mathrm{between}\mathrm{these}$$$$\mathrm{sides}\mathrm{is}\mathrm{either}0°\mathrm{or}180°.$$$$\mathrm{The}\mathrm{bisector}\mathrm{of}\mathrm{the}anglec\mathrm{cannot}\mathrm{be}$$$$\mathrm{perpendicular}\mathrm{to}\mathrm{the}\mathrm{side}b\mathrm{because}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}$$$$\mathrm{angle}\mathrm{between}\mathrm{the}\mathrm{sides}a\mathrm{and}b.$$

Commented by Frix last updated on 16/Sep/24

$$\delta =\sqrt{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}$$$$\tan \alpha =\frac{\delta }{b^2+c^2-a^2}$$$$\tan \beta =\frac{\delta }{a^2+c^2-b^2}$$$$\tan \gamma =\frac{\delta }{a^2+b^2-c^2}$$$$2\tan \gamma +\tan \alpha =0$$$$\iff$$$$a^2-3b^2-c^2=0$$$$\mathrm{Let}c=1\Rightarrow a=\sqrt{3b^2+1}\wedge \delta =2b\sqrt{1-b^2}$$$$\Rightarrow 0<b<1$$$$\tan \alpha =-\frac{\sqrt{1-b^2}}{b}$$$$\tan \beta =\frac{b\sqrt{1-b^2}}{b^2+1}$$$$\tan \gamma =\frac{\sqrt{1-b^2}}{2b}$$$$\mathrm{You}\mathrm{can}\mathrm{draw}\mathrm{any}\mathrm{of}\mathrm{these}\mathrm{to}\mathrm{see}{\mathrm{what}}^{\prime }\mathrm{s}$$$$\mathrm{perpendicular}\mathrm{or}\mathrm{not}...$$

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