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ini_mini
Pe: 2011-02-22T12:52:49+02:00 In: In:

geometrie a VII-a

ABC triunghi
< A = 90 grade
AD perpendicular pe BC
D apartine lui BC
Aratati ca 1/AD la a doua = 1/AB la adoua + 1/AC la a doua.

Va rog foarte mult sa ma ajutati.

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2 raspunsuri

  1. yuxdar

    \rm{AD=\frac{AB\cdot AC}{BC}\Rightarrow AD^2=\frac{AB^2\cdot AC^2}{BC^2}\Rightarrow \frac{1}{AD^2}=\frac{BC^2}{AB^2\cdot AC^2} (1)\\Th.Pitagora \Rightarrow BC^2=AB^2+AC^2 (2)\\(1), (2)\Rightarrow \frac{1}{AD^2}=\frac{AB^2+AC^2}{AB^2\cdot AC^2}=\frac{AB^2}{AB^2AC^2} +\frac{AC^2}{AB^2AC^2}=\frac{1}{AC^2}+\frac{1}{AB^2}=\frac{1}{AB^2}+\frac{1}{AC^2}.\\Deci: \frac{1}{AD^2}=\frac{1}{AB^2}+\frac{1}{AC^2.}}

  2. ini_mini

    yuxdar wrote: \rm{AD=\frac{AB\cdot AC}{BC}\Rightarrow AD^2=\frac{AB^2\cdot AC^2}{BC^2}\Rightarrow \frac{1}{AD^2}=\frac{BC^2}{AB^2\cdot AC^2} (1)\\Th.Pitagora \Rightarrow BC^2=AB^2+AC^2 (2)\\(1), (2)\Rightarrow \frac{1}{AD^2}=\frac{AB^2+AC^2}{AB^2\cdot AC^2}=\frac{AB^2}{AB^2AC^2} +\frac{AC^2}{AB^2AC^2}=\frac{1}{AC^2}+\frac{1}{AB^2}=\frac{1}{AB^2}+\frac{1}{AC^2}.\\Deci: \frac{1}{AD^2}=\frac{1}{AB^2}+\frac{1}{AC^2.}}

    Va multumesc mult..urmarind problema acum nu mi se mai pare asa de grea,dar din pacate nu m-am gandit la abordarea asta.de fapt nu prea stiam cu ce sa incep 🙁 Inca o data multumesc,va doresc o seara linistita!

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