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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

$$provethatthelengthesoftheblue$$$$linesare4timesofthe$$$$correspondingsidelengthesofthe$$$$triangle.$$$$allfigureswhichlooklikesquares$$$$aresquares.$$

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

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

$$p^2=b^2+c^2-2bc\cos (180°-\alpha )$$$$=b^2+c^2+2bc\cos \alpha$$$$=b^2+c^2+(b^2+c^2-a^2)$$$$=2(b^2+c^2)-a^2=4m_a^2$$$$\Rightarrow p=2m_{a}with{m}_a=mediantosidea$$$$\frac{\sin {\alpha }_1}{c}=\frac{\sin {\alpha }_2}{b}=\frac{\sin \alpha }{p}$$$$\sin {\alpha }_1=\frac{c\sin \alpha }{2m_a}=\frac{ac}{2m_a}\times \frac{\sin \alpha }{a}$$$$=\frac{ac}{2m_a}\times \frac{\sin \gamma }{c}=\frac{a}{2}\times \frac{\sin \gamma }{m_a}$$$$\Rightarrow thatmeansDliesonthemedian,$$$$i.e.Disthemediancenter\left(centoid\right).$$$$DA=\frac{2m_a}{3}=\frac{p}{3}$$$${AA}^{\prime }=p$$$${DA}^{\prime }=\frac{p}{3}+p=\frac{4p}{3}$$$$similarly$$$${DB}^{\prime }=\frac{4q}{3}$$$${DC}^{\prime }=\frac{4r}{3}$$$$\mathrm{\Delta }{A}^{\prime }{B}^{\prime }{C}^{\prime }\sim \mathrm{\Delta }ABC$$$$\frac{B^{\prime }{C}^{\prime }}{BC}=\frac{{DB}^{\prime }}{DB}=4$$$$B^{\prime }{C}^{\prime }=4aand{B}^{\prime }{C}^{\prime }//BC$$$$similarly$$$$A^{\prime }{C}^{\prime }=4band{A}^{\prime }{C}^{\prime }//AC$$$$A^{\prime }{B}^{\prime }=4cand{A}^{\prime }{B}^{\prime }//AB$$

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