a: Xét ΔADC có OE//DC
nên \(\dfrac{OE}{DC}=\dfrac{AO}{AC}\left(1\right)\)
Xét ΔBDC có OF//DC
nên \(\dfrac{OF}{DC}=\dfrac{BO}{BD}\left(2\right)\)
Xét ΔOAB và ΔOCD có
\(\widehat{OAB}=\widehat{OCD}\)(hai góc so le trong, AB//CD)
\(\widehat{AOB}=\widehat{COD}\)(hai góc đối đỉnh)
Do đó: ΔOAB~ΔOCD
=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}\)
=>\(\dfrac{OC}{OA}=\dfrac{OD}{OB}\)
=>\(\dfrac{OC+OA}{OA}=\dfrac{OD+OB}{OB}\)
=>\(\dfrac{AC}{AO}=\dfrac{BD}{BO}\)
=>\(\dfrac{AO}{AC}=\dfrac{BO}{BD}\left(3\right)\)
Từ (1),(2),(3) suy ra OE=OF
b: Xét ΔCAB có OF//AB
nên \(\dfrac{OF}{AB}=\dfrac{CO}{CA}\)
\(\dfrac{OE}{DC}+\dfrac{OF}{AB}=\dfrac{AO}{AC}+\dfrac{CO}{CA}=\dfrac{AC}{AC}=1\)
=>\(\dfrac{OE}{AB}+\dfrac{OE}{CD}=1\)
=>\(OE\left(\dfrac{1}{AB}+\dfrac{1}{CD}\right)=1\)
=>\(\dfrac{1}{AB}+\dfrac{1}{CD}=\dfrac{1}{OE}\)
