Problema (nível 4): Na figura seguinte os arcos $\stackrel{\;\frown}{AB}$ e $\stackrel{\;\frown}{BD}$ têm o mesmo comprimento e $M$ é o pé da perpendicular traçada a partir de $B$ sobre o segmento de recta $[AC]$. (image1) Prova que $\overline{AM}=\overline{CD}+\overline{CM}$.


Conteúdo referenciado:
[image1]
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Resolução: Tracemos o segmento $[BE]$ paralelo a $[AC]$ e seja $N$ o pé da perpendicular traçada a partir de $E$ sobre o segmento de recta $[AC]$: (image2) Então $\overline{BE}=\overline{MN}$ e $\overline{CM}=\overline{AN}$. Como os comprimentos de $\stackrel{\;\frown}{AB}$ e $\stackrel{\;\frown}{BD}$ são iguais, o mesmo se passando com $\stackrel{\;\frown}{AE}$ e $\stackrel{\;\frown}{BC}$, os arcos $\stackrel{\;\frown}{BE}$ e $\stackrel{\;\frown}{DC}$ têm o mesmo comprimento. Portanto, as respectivas cordas têm comprimentos iguais, isto é, $\overline{CD}=\overline{BE}=\overline{MN}$. Logo $\overline{AM}=\overline{AN}+\overline{NM}=\overline{CM}+\overline{CD }$.


Conteúdo referenciado:
[image2]
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