Trapez sau paralelogram
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Trapez sau paralelogram
Se noteaza cu \( I \) centrul unui cerc inscris intr-un patrulater \( ABCD \). Fie \( M \) si \( N \) mijloacele laturilor \( AB \), respectiv \( CD \). Daca \( \frac{IM}{AB}=\frac{IN}{CD} \), atunci sa se demonstreze ca \( ABCD \) este trapez sau paralelogam.
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Claudiu Mindrila
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Scriind relatia medianei si folosind ipoteza obtinem:
\( \frac{IM^{2}}{AB^{2}}=\frac{IN^{2}}{CD^{2}}\Longrightarrow\frac{AB^{2}}{CD^{2}}=\frac{IM^{2}}{IN^{2}}=\frac{2\left(IA^{2}+IB^{2}\right)-AB^{2}}{2\left(IC^{2}+ID^{2}\right)-CD^{2}}\Longrightarrow\frac{AB^{2}}{IA^{2}+IB^{2}}=\frac{CD^{2}}{IC^{2}+ID^{2}} \). Folosind teorema cosinusului avem:
\( 1+\frac{2\cdot IA\cdot IB\cdot\cos\widehat{AIB}}{IA^{2}+IB^{2}}=1+\frac{2\cdot IC\cdot ID\cdot\cos\widehat{CID}}{IC^{2}+ID^{2}} \) adica \( \frac{IA\cdot IB\cdot\cos\widehat{AIB}}{IA^{2}+IB^{2}}=\frac{IC\cdot ID\cdot\cos\widehat{CID}}{IC^{2}+ID^{2}} \).
Dar \( \widehat{AIB}+\widehat{CID}=\left(180^{\circ}-\frac{\widehat{A}}{2}-\frac{\widehat{B}}{2}\right)+\left(180^{\circ}-\frac{\widehat{C}}{2}-\frac{\widehat{D}}{2}\right)=180^{\circ} \)
Daca \( \widehat{AIB}\neq\widehat{CID} \) sa presupunem ca \( \widehat{AIB}>\widehat{CID}\Longrightarrow\widehat{AIB}>\frac{\widehat{AIB}+\widehat{CID}}{2}=90>\widehat{CID} \) prin urmare \( \cos\widehat{AIB} \) va fi negativ, ceea ce este o contradictie. Asadar \( \widehat{AIB}=\widehat{CID}\Longrightarrow\widehat{AIB}=90^{\circ}\Longrightarrow\widehat{A}+\widehat{B}=180^{\circ}\Longrightarrow AB\parallel CD \)
\( \frac{IM^{2}}{AB^{2}}=\frac{IN^{2}}{CD^{2}}\Longrightarrow\frac{AB^{2}}{CD^{2}}=\frac{IM^{2}}{IN^{2}}=\frac{2\left(IA^{2}+IB^{2}\right)-AB^{2}}{2\left(IC^{2}+ID^{2}\right)-CD^{2}}\Longrightarrow\frac{AB^{2}}{IA^{2}+IB^{2}}=\frac{CD^{2}}{IC^{2}+ID^{2}} \). Folosind teorema cosinusului avem:
\( 1+\frac{2\cdot IA\cdot IB\cdot\cos\widehat{AIB}}{IA^{2}+IB^{2}}=1+\frac{2\cdot IC\cdot ID\cdot\cos\widehat{CID}}{IC^{2}+ID^{2}} \) adica \( \frac{IA\cdot IB\cdot\cos\widehat{AIB}}{IA^{2}+IB^{2}}=\frac{IC\cdot ID\cdot\cos\widehat{CID}}{IC^{2}+ID^{2}} \).
Dar \( \widehat{AIB}+\widehat{CID}=\left(180^{\circ}-\frac{\widehat{A}}{2}-\frac{\widehat{B}}{2}\right)+\left(180^{\circ}-\frac{\widehat{C}}{2}-\frac{\widehat{D}}{2}\right)=180^{\circ} \)
Daca \( \widehat{AIB}\neq\widehat{CID} \) sa presupunem ca \( \widehat{AIB}>\widehat{CID}\Longrightarrow\widehat{AIB}>\frac{\widehat{AIB}+\widehat{CID}}{2}=90>\widehat{CID} \) prin urmare \( \cos\widehat{AIB} \) va fi negativ, ceea ce este o contradictie. Asadar \( \widehat{AIB}=\widehat{CID}\Longrightarrow\widehat{AIB}=90^{\circ}\Longrightarrow\widehat{A}+\widehat{B}=180^{\circ}\Longrightarrow AB\parallel CD \)
elev, clasa a X-a, C. N. "C-tin Carabella", Targoviste