Fie triunghiul ascutitunghic \( ABC \) cu ortocentrul \( H \), incercul \( w = C(I,r) \) si circumcercul \( C(O) \).
Notam punctul \( D \) unde cercul \( w \) atinge dreapta \( BC \). Sa se arate ca \( AH = r\ \Longleftrightarrow\ OI = OD \).
AH=r <=> OD=OI
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Marius Mainea
- Gauss
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Folosim relatiile:
\( AH=2R\cos A \), \( OI^2=R^2-2Rr \), \( BD\cdot DC=R^2-OD^2, r=4r\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}, BD=p-b, DC=p-c \).
Avem \( AH=r\Longleftrightarrow 2R\cos A=r \Longleftrightarrow \cos B+\cos C - \cos A=1 \).
Deasemenea \( OI=OD \Longleftrightarrow (p-b)(p-c)=2Rr\Longleftrightarrow \cos B+\cos C-\cos A=1 \).
\( AH=2R\cos A \), \( OI^2=R^2-2Rr \), \( BD\cdot DC=R^2-OD^2, r=4r\sin{\frac{A}{2}}\sin{\frac{B}{2}}\sin{\frac{C}{2}}, BD=p-b, DC=p-c \).
Avem \( AH=r\Longleftrightarrow 2R\cos A=r \Longleftrightarrow \cos B+\cos C - \cos A=1 \).
Deasemenea \( OI=OD \Longleftrightarrow (p-b)(p-c)=2Rr\Longleftrightarrow \cos B+\cos C-\cos A=1 \).