CMR BD vuông góc với AM

E

erokl1

Nè bạn

Ta có: $AM vuông góc BD$
\Leftrightarrow $\overrightarrow{BD}$.$\overrightarrow{AM}$ =0
\Leftrightarrow $\overrightarrow{BA}$.$\overrightarrow{AM}$ + $\overrightarrow{BC}$.$\overrightarrow{AM}$ =0
\Leftrightarrow $-AB^2$ + $\overrightarrow{BA}$.$\overrightarrow{BM}$ +$\overrightarrow{BC}$.$\overrightarrow{AB}$ +$\overrightarrow{BC}$.$\overrightarrow{BM}$ = 0
\Leftrightarrow $-AB^2$ +$ BA.BM.cosB $-$3BM.AB.cosB$ + 3$BM^2$
\Leftrightarrow $-AB^2$ - $2BA.BM.cosB$ + $3BM^2$ = 0
\Leftrightarrow $-AB^2$ - 2BA.BM.$\frac{AB^2+BC^2-AC^2}{2AB.BC}$ + $3BM^2$ = 0
\Leftrightarrow $-AB^2$ - $\frac{AB^2+9BM^2-4AB^2}{3}$ + $3BM^2$ = 0
\Leftrightarrow 0 = 0 , luôn đúng. Vậy AM vuông góc BD
 
You must log in or register to reply here.
Top Bottom