a, Ta có: \(\left\{{}\begin{matrix}OM//CD\\ON//CD\\ON//AB\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{OM}{CD}=\frac{OA}{AC}\left(1\right)\\\frac{ON}{CD}=\frac{BN}{BC}\left(2\right)\\\frac{OA}{AC}=\frac{BN}{BC}\left(3\right)\end{matrix}\right.\)
Từ: \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\frac{OM}{CD}=\frac{ON}{CD}\Rightarrow OM=ON\left(đpcm\right)\)
b, Ta có: \(OM//AB\Rightarrow\frac{OM}{AB}=\frac{DM}{DA}\left(4\right)\)
Và: \(\) \(OM//CD\Rightarrow\frac{OM}{CD}=\frac{AM}{AD}\left(5\right)\)
Từ: \(\left(4\right)\left(5\right)\Rightarrow\frac{OM}{AB}+\frac{OM}{CD}=\frac{DM+AM}{DA}=1\)
Chia 2 vế cho \(OM\) ta được: \(\frac{1}{AB}+\frac{1}{CD}=\frac{1}{OM}\)
Hay: \(\frac{1}{AB}+\frac{1}{CD}=\frac{2}{MN}\)
Tự vẽ hình
a.Vì MO//CD (MN//CD,O\(\in\) MN) \(\Rightarrow\) \(\frac{AO}{AC}=\frac{OM}{CD}\) (1)
Vì NO//CD (MN//CD,O\(\in\) MN) \(\Rightarrow\frac{ON}{CD}=\frac{BO}{AD}\) (2)
Vì AB//CD \(\Rightarrow\frac{AO}{AC}=\frac{BO}{BD}\) (3)
Từ (1),(2),(3) \(\Rightarrow\frac{OM}{CD}=\frac{ON}{CD}\Rightarrow OM=ON\left(đpcm\right)\)
b.
\(\frac{1}{AB}+\frac{1}{CD}=\frac{2}{MN}\\ \Leftrightarrow\frac{MN}{AB}+\frac{MN}{CD}=2\\ \Leftrightarrow\frac{MO}{AB}+\frac{ON}{AB}+\frac{MO}{CD}+\frac{ON}{CD}=2\\ \Leftrightarrow\frac{MO}{AB}+\frac{ON}{CD}+\frac{MO}{CD}+\frac{ON}{AB}=2\)
Dễ có: \(\frac{MO}{AB}=\frac{DO}{DB};\frac{ON}{CD}=\frac{BO}{DB};\frac{MO}{CD}=\frac{AO}{AC};\frac{ON}{AB}=\frac{CO}{AC}\) (Sử dụng hệ quả của định lí Thales)
Thay vào ta có được:
\(\Leftrightarrow\frac{DO}{DB}+\frac{BO}{DB}+\frac{AO}{AC}+\frac{CO}{AC}=2\\ \Leftrightarrow\frac{DB}{DB}+\frac{AC}{AC}=2\\ \Leftrightarrow1+1=2\left(đpcm\right)\)
