mateforum.ro

Fie \( ABC \) un triunghi si \( M \) un punct pe latura \( BC \) astfel incat \( BM=3CM \). Stiind ca \( m(\angle AMB)=\frac{1}{2}m(\angle B) \) sa se arate ca \( a=2(b-c) \).

Virgil Nicula wrote: Lema. Sa se arate ca in \( \triangle ABC \) avem \( B=2\cdot C\ \Longleftrightarrow \ a=c\ \ \vee\ \ b^2=c(c+a)\ . \)

Virgil Nicula wrote: Fie \( \triangle ABC \) si \( M\in (BC) \) astfel incat \( m(\angle AMB)=\frac B2 \) . Sa se arate ca

\( MB=c \) sau \( a\cdot MC+c(a+c)=b^2 \) . In particular, \( MC=\frac a4\ \Longleftrightarrow\ a=2(b-c) \) .

Dem. Aplicam relatia Stewart cevienei \( [AM \) in \( \triangle ABC \) : \( AM^2\cdot a+MB\cdot MC\cdot a=c^2\cdot MC+b^2\cdot MB\ (*)\ . \)

Folosind lema precedenta in \( \triangle ABM \) si se obtine \( MB=c\ \ \vee\ \ AM^2=c(c+MB)\ . \)

Daca \( AM^2=c(c+MB) \) , atunci tinand seama de relatia \( (*) \) obtinem \( ac^2+ac(a-MC)+a(a-MC)MC= \)

\( c^2MC+b^2(a-MC)\Longleftrightarrow a\cdot MC^2+\left[\left(c^2+ac-b^2\right)-a^2\right]\cdot MC-a\left(c^2+ac-b^2\right)=0\Longleftrightarrow \)

\( \left(MC-a\right)\left[aMC+c(a+c)-b^2\right]=0\Longleftrightarrow \) \( MC=a\ \ \vee\ \ a\cdot MC+c(a+c)=b^2\Longleftrightarrow \) \( a\cdot MC+c(a+c)=b^2\ . \)