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

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

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

Commented by Frix last updated on 16/Sep/24

If the bisector of a side is perpendicular to the other side ⇒ the angle between these sides is either 0° or 180°. The bisector of the angle c cannot be perpendicular to the side b because it′s the angle between the sides a and b.

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

Commented by Frix last updated on 16/Sep/24

δ=(√((a+b+c)(a+b−c)(a+c−b)(b+c−a))) tan α =(δ/(b^2 +c^2 −a^2 )) tan β =(δ/(a^2 +c^2 −b^2 )) tan γ =(δ/(a^2 +b^2 −c^2 )) 2tan γ +tan α =0 ⇔ a^2 −3b^2 −c^2 =0 Let c=1 ⇒ a=(√(3b^2 +1))∧δ=2b(√(1−b^2 )) ⇒ 0<b<1 tan α =−((√(1−b^2 ))/b) tan β =((b(√(1−b^2 )))/(b^2 +1)) tan γ =((√(1−b^2 ))/(2b)) You can draw any of these to see what′s perpendicular or not...

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

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