Question

Multumesc!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

$\Delta ABC \equiv \Delta C'B'C \Rightarrow B'C=a,~B'C'=b~si~CC'=c~; \\ \\ m(\ \textless \ C'CB)=m(\ \textless \ ABC),~iar~cum~\ \textless \ ABC~si~\ \textless \ ACB~sunt \\ \\ complementare,~rezulta~\ \textless \ C'CB~si~\ \textless \ ACB ~sunt~complementare, \\ \\ adica~m(\ \textless \ C'CB)+m(\ \textless \ ACB)=90 \textdegree \Rightarrow m(\ \textless \ BCB')=180 \textdegree-90 \textdegree= \\ \\ =90 \textdegree.$

$A_{ABB'C'}= \frac{(B'C'+AB) \cdot AC'}{2}= \frac{(b+c)(b+c)}{2}= \frac{(b+c)^2}{2} . \\ \\ A_{ABC}= \frac{AB \cdot AC}{2}= \frac{bc}{2}~~;~~ A_{B'C'C}= \frac{B'C' \cdot CC'}{2}= \frac{bc}{2} ~~;~~ \\ \\ A_{BCB'}= \frac{BC \cdot B'C}{2}= \frac{a \cdot a}{2}= \frac{a^2}{2} . \\ \\ A_{ABB'C'}=A_{ABC}+A_{B'C'C}+A_{BCB'}= \frac{bc}{2}+ \frac{bc}{2}+ \frac{a^2}{2} =bc+ \frac{a^2}{2} .$

$Egalam:~ \frac{(b+c)^2}{2}= bc +\frac{a^2}{2} \Leftrightarrow \frac{b^2+2bc+c^2}{2}= bc+\frac{a^2}{2} \Leftrightarrow \\ \\ \Leftrightarrow \frac{b^2+c^2}{2}+bc= bc+ \frac{a^2}{2} \Leftrightarrow \boxed{b^2+c^2=a^2} -~si~astfel~am~demonstrat \\ \\ teorema~lui~Pitagora.$