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

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

Ta có: $\widehat{AMB}+\widehat{AMC}={180}^0$

$\widehat{AMC}={180}^0-\widehat{AMB}$

$cos\widehat{AMB}=-cos\left({180}^0-\widehat{AMB}\right)=-cos\widehat{AMC}$

$\Rightarrow cos\widehat{AMB}+cos\widehat{AMC}=-cos\widehat{AMC}+cos\widehat{AMC}=0$

b) Xét ΔAMB, ta có:

AB² = MA² + MB² – 2MA.MB.cos$\widehat{AMB}$ 

⇔ MA² + MB² – AB² = 2MA.MB.cos $\widehat{AMB}$(1)

Xét ΔAMC, ta có:

AC² = MA² + MC² – 2MA.MC.cos $\widehat{AMC}$

⇔ MA² + MC² – AC² = 2MA.MC.cos $\widehat{AMC}$ (2)

c) Cộng vế với vế của (1) với (2), ta được:

MA² + MB² – AB² + MA² + MC² – AC²

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

$$\begin{gathered} \iff 2M{A}^2+\frac{B{C}^2}{4}–A{B}^2+\frac{B{C}^2}{4}–A{C}^2 \\ =2MA.\frac{BC}{2}.cos\widehat{AMB} +2MA.\frac{BC}{2}.cos\widehat{AMC} \end{gathered}$$

(Vì $MB=MC=\frac{BC}{2}$)

$\iff 2M{A}^2+\frac{B{C}^2}{2}–A{B}^2–A{C}^2=2MA.\frac{BC}{2}.\left(cos\widehat{AMB} +cos\widehat{AMC}\right)$

$\iff 2M{A}^2+\frac{B{C}^2}{2}–A{B}^2–A{C}^2=\text{ 0}$ (vì $cos\widehat{AMB} +cos\widehat{AMC}=0$)

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

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