Question

In trapezul AMCD din imagine demonatrati ca dreptele DM si CM sunt perpendiculare.

Answer 1

Daca demonstram ca DC² =  DM² + MC²   inseamna ca DC este ipotenuza triunghiului DMC -> DMC triunghi dreptunghic -> m(DMC) = 90

Mai intai aflam MC, notam cu H proiectia perpendiculara a lui M pe DC.

$In \triangle_{HMC}: \ MH = 3, CH = 6,25 - 4 = 2,25 = \frac{225}{100}= \frac{45}{20} \Rightarrow CH= \frac{9}{4} \\ \\ CM^2 = CH^2 + MH^2 \Rightarrow CM^2 = (\frac{9}{4})^2 + 3^2 \Rightarrow CM^2 = \frac{81}{16}+9 \Rightarrow \\ \\ \Rightarrow CM^2 = \frac{81+144}{16} \Rightarrow CM^2 = \frac{225}{16} \Rightarrow CM = \frac{15}{4} \\ \\ \ Acum, in \triangle_{DMA}: \\ \\ DM^2 = DA^2+AM^2 = 3^2+4^2 = 9 +16 = 25 \Rightarrow DM = 5$

$\ Acum putem calcula in sfarsit ipotenuza in \triangle_{DMC}: \\ \\ DC = 6,25 \Rightarrow DC = \frac{625}{100}\Rightarrow DC = \frac{25}{4} \\ \\ DC^2 = DM^2+MC^2 \Rightarrow DC^2 = 5^2 + (\frac{15}{4})^2 =25 + \frac{225}{16} = \frac{400+225}{16} = \frac{625}{16} \\ \\ \Rightarrow DC^2 = \frac{625}{16} \Rightarrow DC = \frac{25}{4} (A) \rightarrow \triangle_{DMC} - \triangle dreptunghic ,\\ \\ \ cu DC ipotenuza \rightarrow DM\perp CM$