Giải bài 3.16 trang 44

Đề bài

Cho tam giác ABC có trung tuyến AM. Chứng minh rằng:

a) $\cos \widehat{AMB}+\cos \widehat{AMC}=0$;

b) $M{A}^2+M{B}^2-A{B}^2=2.MA.MB.\cos \widehat{AMB}$ và $M{A}^2+M{C}^2-A{C}^2=2.MA.MC.\cos \widehat{AMC}$;

c) $M{A}^2=\frac{2\left(A{B}^2+A{C}^2\right)-B{C}^2}{4}$ (công thức đường trung tuyến).

Lời giải chi tiết

 

a) Ta có: $\widehat{AMB}+\widehat{AMC}={180}^o$

$\Rightarrow \cos \widehat{AMB}=-\cos \widehat{AMC}$

hay $\cos \widehat{AMB}+\cos \widehat{AMC}=0$.

b) Áp dụng định lí cos trong tam giác AMB ta có:

$A{B}^2=M{A}^2+M{B}^2-2MA.MB\cos \widehat{AMB}$

$\iff M{A}^2+M{B}^2-A{B}^2=2MA.MB\cos \widehat{AMB}$ (1)

Tương tự, áp dụng định lí cos trong tam giác AMB ta được:

$A{C}^2=M{A}^2+M{C}^2-2MA.MC\cos \widehat{AMC}$

$\iff M{A}^2+M{C}^2-A{C}^2=2MA.MC\cos \widehat{AMC}$ (2)

c) Từ (1), suy ra $M{A}^2=A{B}^2-M{B}^2+2MA.MB\cos \widehat{AMB}$.

Từ (2), suy ra $M{A}^2=A{C}^2-M{C}^2+2MA.MC\cos \widehat{AMC}$.

Cộng vế với vế ta được:

$2M{A}^2=\left(A{B}^2-M{B}^2+2MA.MB\cos \widehat{AMB}\right)$

$+\left(A{C}^2-M{C}^2+2MA.MC\cos \widehat{AMC}\right)$

$\iff 2M{A}^2=A{B}^2+A{C}^2-M{B}^2-M{C}^2$

$+2MA.MB\cos \widehat{AMB}+2MA.MC\cos \widehat{AMC}$.

Mà: $MB=MC=\frac{BC}{2}$ (do AM là trung tuyến)

$\Rightarrow 2M{A}^2=A{B}^2+A{C}^2-{\left(\frac{BC}{2}\right)}^2-{\left(\frac{BC}{2}\right)}^2$

$+2MA.MB\cos \widehat{AMB}+2MA.MB\cos \widehat{AMC}$

$\iff 2M{A}^2=A{B}^2+A{C}^2-2.{\left(\frac{BC}{2}\right)}^2$

$+2MA.MB\left(\cos \widehat{AMB}+\cos \widehat{AMC}\right)$

$\iff 2M{A}^2=A{B}^2+A{C}^2-{\frac{BC}{2}}^2$

$\iff M{A}^2=\frac{A{B}^2+A{C}^2-{\frac{BC}{2}}^2}{2}$

$\iff M{A}^2=\frac{2\left(A{B}^2+A{C}^2\right)-B{C}^2}{4}$.

Cách 2:

Theo ý a, ta có: $\cos \widehat{AMC}=-\cos \widehat{AMB}$

Từ đẳng thức (1): suy ra $\cos \widehat{AMB}=\frac{M{A}^2+M{B}^2-A{B}^2}{2.MA.MB}$

$\Rightarrow \cos \widehat{AMC}=-\cos \widehat{AMB}=-\frac{M{A}^2+M{B}^2-A{B}^2}{2.MA.MB}$.

Thế $\cos \widehat{AMC}$vào biểu thức (2), ta được:

$M{A}^2+M{C}^2-A{C}^2=2MA.MC.\left(-\frac{M{A}^2+M{B}^2-A{B}^2}{2.MA.MB}\right)$

Lại có: $MB=MC=\frac{BC}{2}$ (do AM là trung tuyến)

$\Rightarrow M{A}^2+{\left(\frac{BC}{2}\right)}^2-A{C}^2=2MA.MB.\left(-\frac{M{A}^2+M{B}^2-A{B}^2}{2.MA.MB}\right)$

$\iff M{A}^2+{\left(\frac{BC}{2}\right)}^2-A{C}^2=-\left(M{A}^2+M{B}^2-A{B}^2\right)$

$\iff M{A}^2+{\left(\frac{BC}{2}\right)}^2-A{C}^2+M{A}^2+{\left(\frac{BC}{2}\right)}^2-A{B}^2=0$

$\iff 2M{A}^2-A{B}^2-A{C}^2+{\frac{BC}{2}}^2=0$

$\iff 2M{A}^2=A{B}^2+A{C}^2-{\frac{BC}{2}}^2$

$\iff M{A}^2=\frac{A{B}^2+A{C}^2-{\frac{BC}{2}}^2}{2}$

$\iff M{A}^2=\frac{2\left(A{B}^2+A{C}^2\right)-B{C}^2}{4}$.