Bài tập Chứng minh đẳng thức về tích vô hướng của vectơ hoặc về độ dài đoạn thẳng (có lời giải)

 $\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MB}.\overrightarrow{MC}+\overrightarrow{MC}.\overrightarrow{MA}=\frac{{a}^{2}}{4}$

 $\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MB}.\overrightarrow{MC}+\overrightarrow{MC}.\overrightarrow{MA}={a}^{2}$

 $\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MB}.\overrightarrow{MC}+\overrightarrow{MC}.\overrightarrow{MA}=2{a}^{2}$

 $\overrightarrow{MA}.\overrightarrow{MB}+\overrightarrow{MB}.\overrightarrow{MC}+\overrightarrow{MC}.\overrightarrow{MA}=\frac{{a}^{2}}{2}$

 $\overrightarrow{MH}.\overrightarrow{MA}=\frac{1}{2}B{C}^{2}$

 $\overrightarrow{MH}.\overrightarrow{MA}=\frac{1}{4}B{C}^{2}$

 $\overrightarrow{MH}.\overrightarrow{MA}=\frac{3}{4}B{C}^{2}$

 $\overrightarrow{MH}.\overrightarrow{MA}=\frac{5}{4}B{C}^{2}$

 ${c}^{2}.C{M}^{2}={a}^{2}.A{M}^{2}+{b}^{2}.B{M}^{2}+\left({a}^{2}+{b}^{2}-{c}^{2}\right).AM.BM$

 ${c}^{2}.C{M}^{2}={a}^{2}.A{M}^{2}+{b}^{2}.B{M}^{2}+\left({a}^{2}-{b}^{2}-{c}^{2}\right).AM.BM$

 ${c}^{2}.C{M}^{2}={a}^{2}.A{M}^{2}-{b}^{2}.B{M}^{2}+\left({a}^{2}+{b}^{2}-{c}^{2}\right).AM.BM$

 ${c}^{2}.C{M}^{2}={a}^{2}.A{M}^{2}+{b}^{2}.B{M}^{2}-\left({a}^{2}+{b}^{2}-{c}^{2}\right).AM.BM$

 $\overrightarrow{AM}.\overrightarrow{AN}={d}^{2}+{R}^{2}$

 $\overrightarrow{AM}.\overrightarrow{AN}={d}^{2}-{R}^{2}$

 $\overrightarrow{AM}.\overrightarrow{AN}=2{d}^{2}+{R}^{2}$

 $\overrightarrow{AM}.\overrightarrow{AN}={d}^{2}+2{R}^{2}$

 $\overrightarrow{AE}.\overrightarrow{AC}+\overrightarrow{BE}.\overrightarrow{BD}=A{B}^{2}$

 $\overrightarrow{AE}.\overrightarrow{AC}-\overrightarrow{BE}.\overrightarrow{BD}=A{B}^{2}$

 $\overrightarrow{AE}.\overrightarrow{AC}+\overrightarrow{BE}.\overrightarrow{BD}=-A{B}^{2}$

 $\overrightarrow{AE}.\overrightarrow{AC}+\overrightarrow{BE}.\overrightarrow{BD}=2A{B}^{2}$