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Un trapez si o axa radicala (Own).
Posted: Sun Feb 17, 2008 5:05 pm
by Virgil Nicula
Se considera trapezul \( ABCD \), unde \( AB\ \parallel\ CD \). Fiind date punctele \( M,\ N \) care apartin dreptei \( AB \) sa se arate ca \( P\in CM\cap DN \) apartine axei radicale a cercurilor circumscrise triunghiurilor \( ADM \) si \( CBN \).
Posted: Mon Feb 18, 2008 7:46 pm
by mihai++
Cred ca ori imi scapa mie ceva ori problema e incorecta.
\( C(AMD) \) se intersecteaza cu \( C(DNC) \) in doua puncte, unul dintre ele fiind \( D \), care deci este pe axa radicala.
Daca \( P \) ar apartine axei radicale atunci si \( N\in DP \) apartine axei radicale, deci \( N \) are puteri egale fata de cele doua cercuri, dar \( N \) are puterea \( 0 \) fata de \( C(CND) \), deci si fata de \( C(AMD) \), adica \( NM\cdot NA=0 \), ceea ce e fals.
Posted: Mon Feb 18, 2008 8:42 pm
by mihai++
Fie \( T=DN\cap C(AMD) \) si \( S=CM\cap C(CNB) \). \( P\in \) axei radicale \( \Leftrightarrow PT\cdot PD=PS\cdot PC\Leftrightarrow \frac {PT}{PS}=\frac{PC}{PD}=\frac{PM}{PN} \Leftrightarrow TMNS \)-inscriptibil
\( \Leftrightarrow \hat{MTN}=\hat{MSN} \Leftrightarrow 180^{\circ}-\hat{MTN}=180^{\circ}-\hat{MSN} \Leftrightarrow \)
\( \hat{MTD}=\hat{NBC}\Leftrightarrow 180^{\circ}-\hat{MTD}=180^{\circ}-\hat{NBC}\Leftrightarrow \)
\( \hat{A}=\hat{B}\Leftrightarrow ABCD \)-trapez isoscel.
cred ca asta lipseste.