a: Xét ΔOAB và ΔOCD có
\(\hat{OAB}=\hat{OCD}\) (hai góc so le trong, AB//CD)
\(\hat{AOB}=\hat{COD}\) (hai góc đối đỉnh)
Do đó: ΔOAB~ΔOCD
=>\(\frac{OA}{OC}=\frac{OB}{OD}=\frac{AC}{BD}\)
=>\(\frac{OA}{OB}=\frac{OC}{OD}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\frac{OA}{OB}=\frac{OC}{OD}=\frac{OA+OC}{OB+BD}=\frac{AC}{BD}\)
=>\(\frac{AO}{AC}=\frac{BO}{BD}\) (1)
Xét ΔADC có OM//DC
nên \(\frac{AO}{AC}=\frac{MO}{DC}\left(2\right)\)
Xét ΔBDC có ON//DC
nên \(\frac{ON}{DC}=\frac{BO}{BD}\left(3\right)\)
Từ (1),(2),(3) suy ra OM=ON
b: Xét ΔDAB có MO//AB
nên \(\frac{MO}{AB}=\frac{DO}{DB}\)
Ta có: \(\frac{MO}{AB}+\frac{NO}{DC}=\frac{DO}{DB}+\frac{OB}{DB}=1\)
=>\(\frac{MO}{AB}+\frac{MO}{DC}=1\)
=>\(MO\left(\frac{1}{AB}+\frac{1}{DC}\right)=1\)
=>\(\frac{1}{AB}+\frac{1}{CD}=\frac{1}{OM}=\frac{2}{M\mathbb{N}}\)
c: ΔOAB~ΔOCD
=>\(\frac{S_{AOB}}{S_{COD}}=\left(\frac{OA}{OC}\right)^2=\left(\frac{OB}{OD}\right)^2\)
=>\(\left(\frac{OA}{OC}\right)^2=\left(\frac{OB}{OD}\right)^2=\frac{a^2}{b^2}=\left(\frac{a}{b}\right)^2\)
=>\(\frac{OA}{OC}=\frac{OB}{OD}=\frac{a}{b}\)
Ta có: \(\frac{OA}{OC}=\frac{a}{b}\)
=>\(\frac{S_{AOB}}{S_{BOC}}=\frac{a}{b}\)
=>\(\frac{a^2}{S_{BOC}}=\frac{a}{b}=\frac{a^2}{ab}\)
=>\(S_{BOC}=ab\)
Ta có: \(\frac{OB}{OD}=\frac{a}{b}\)
=>\(\frac{S_{AOB}}{S_{AOD}}=\frac{a}{b}\)
=>\(\frac{a^2}{S_{AOD}}=\frac{a}{b}=\frac{a^2}{ab}\)
=>\(S_{AOD}=ab\)
\(S_{ABCD}=S_{OAB}+S_{OBC}+S_{OAD}+S_{ODC}\)
\(=a^2+ab+b^2+ab=\left(a+b\right)^2\)
