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Question Number 180882 by mr W last updated on 18/Nov/22

Commented by mr W last updated on 18/Nov/22

prove that the lengthes of the blue lines are 4 times of the corresponding side lengthes of the triangle. all figures which look like squares are squares.

$${prove}\:{that}\:{the}\:{lengthes}\:{of}\:{the}\:{blue} \\ $$$${lines}\:{are}\:\mathrm{4}\:{times}\:{of}\:{the} \\ $$$${corresponding}\:{side}\:{lengthes}\:{of}\:{the} \\ $$$${triangle}. \\ $$$${all}\:{figures}\:{which}\:{look}\:{like}\:{squares} \\ $$$${are}\:{squares}. \\ $$

Answered by mr W last updated on 19/Nov/22

Commented by mr W last updated on 19/Nov/22

p^2 =b^2 +c^2 −2bc cos (180°−α) =b^2 +c^2 +2bc cos α =b^2 +c^2 +(b^2 +c^2 −a^2 ) =2(b^2 +c^2 )−a^2 =4m_a ^2 ⇒p=2m_a with m_a =median to side a ((sin α_1 )/c)=((sin α_2 )/b)=((sin α)/p) sin α_1 =((c sin α)/(2m_a ))=((ac )/(2m_a ))×((sin α)/a) =((ac)/(2m_a ))×((sin γ)/c)=(a/2)×((sin γ)/m_a ) ⇒that means D lies on the median, i.e. D is the median center (centoid). DA=((2m_a )/3)=(p/3) AA′=p DA′=(p/3)+p=((4p)/3) similarly DB′=((4q)/3) DC′=((4r)/3) ΔA′B′C′∼ΔABC ((B′C′)/(BC))=((DB′)/(DB))=4 B′C′=4a and B′C′//BC similarly A′C′=4b and A′C′//AC A′B′=4c and A′B′//AB

$${p}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\:\left(\mathrm{180}°−\alpha\right) \\ $$$$\:\:\:\:={b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{bc}\:\mathrm{cos}\:\alpha \\ $$$$\:\:\:\:={b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:=\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} =\mathrm{4}{m}_{{a}} ^{\mathrm{2}} \\ $$$$\Rightarrow{p}=\mathrm{2}{m}_{{a}} \:{with}\:{m}_{{a}} ={median}\:{to}\:{side}\:{a} \\ $$$$\frac{\mathrm{sin}\:\alpha_{\mathrm{1}} }{{c}}=\frac{\mathrm{sin}\:\alpha_{\mathrm{2}} }{{b}}=\frac{\mathrm{sin}\:\alpha}{{p}} \\ $$$$\mathrm{sin}\:\alpha_{\mathrm{1}} =\frac{{c}\:\mathrm{sin}\:\alpha}{\mathrm{2}{m}_{{a}} }=\frac{{ac}\:}{\mathrm{2}{m}_{{a}} }×\frac{\mathrm{sin}\:\alpha}{{a}} \\ $$$$\:\:\:\:\:=\frac{{ac}}{\mathrm{2}{m}_{{a}} }×\frac{\mathrm{sin}\:\gamma}{{c}}=\frac{{a}}{\mathrm{2}}×\frac{\mathrm{sin}\:\gamma}{{m}_{{a}} } \\ $$$$\Rightarrow{that}\:{means}\:{D}\:{lies}\:{on}\:{the}\:{median}, \\ $$$${i}.{e}.\:{D}\:{is}\:{the}\:{median}\:{center}\:\left({centoid}\right). \\ $$$${DA}=\frac{\mathrm{2}{m}_{{a}} }{\mathrm{3}}=\frac{{p}}{\mathrm{3}} \\ $$$${AA}'={p} \\ $$$${DA}'=\frac{{p}}{\mathrm{3}}+{p}=\frac{\mathrm{4}{p}}{\mathrm{3}} \\ $$$${similarly} \\ $$$${DB}'=\frac{\mathrm{4}{q}}{\mathrm{3}} \\ $$$${DC}'=\frac{\mathrm{4}{r}}{\mathrm{3}} \\ $$$$\Delta{A}'{B}'{C}'\sim\Delta{ABC} \\ $$$$\frac{{B}'{C}'}{{BC}}=\frac{{DB}'}{{DB}}=\mathrm{4} \\ $$$${B}'{C}'=\mathrm{4}{a}\:{and}\:{B}'{C}'//{BC} \\ $$$${similarly} \\ $$$${A}'{C}'=\mathrm{4}{b}\:{and}\:{A}'{C}'//{AC} \\ $$$${A}'{B}'=\mathrm{4}{c}\:{and}\:{A}'{B}'//{AB} \\ $$

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