a: 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
=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}\)
=>\(\dfrac{OA}{OB}=\dfrac{OC}{OD}=\dfrac{OA+OC}{OB+OD}=\dfrac{AC}{BD}\)
=>\(\dfrac{AO}{AC}=\dfrac{BO}{BD}\)(2)
Xét ΔADC có OM//DC
nên \(\dfrac{OM}{DC}=\dfrac{AO}{AC}\left(1\right)\)
Xét ΔBDC có ON//DC
nên \(\dfrac{ON}{DC}=\dfrac{BO}{BD}\left(3\right)\)
Từ (1),(2),(3) suy ra OM=ON
b: Xét ΔCAB có ON//AB
nên \(\dfrac{ON}{AB}=\dfrac{CO}{CA}\)
\(\dfrac{ON}{DC}+\dfrac{ON}{AB}=\dfrac{BO}{BD}+\dfrac{CO}{CA}\)
\(=\dfrac{AO}{AC}+\dfrac{CO}{AC}=1\)
=>\(\dfrac{1}{DC}+\dfrac{1}{AB}=\dfrac{1}{ON}=\dfrac{1}{\dfrac{MN}{2}}=\dfrac{2}{MN}\)
c: ΔOAB~ΔOCD
=>\(\dfrac{S_{OAB}}{S_{OCD}}=\left(\dfrac{OA}{OC}\right)^2=\left(\dfrac{OB}{OD}\right)^2=\left(\dfrac{2024}{2025}\right)^2\)
=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}=\dfrac{2024}{2025}\)
OA/OC=2024/2025
=>\(\dfrac{S_{OAB}}{S_{BOC}}=\dfrac{2024}{2025}\)
=>\(S_{BOC}=2025\cdot\dfrac{2024^2}{2024}=2025\cdot2024\left(đvdt\right)\)
\(\dfrac{OA}{OC}=\dfrac{2024}{2025}\)
=>\(\dfrac{S_{AOD}}{S_{DOC}}=\dfrac{2024}{2025}\)
=>\(S_{AOD}=\dfrac{2024}{2025}\cdot2025^2=2024\cdot2025\)
\(S_{ABCD}=S_{AOB}+S_{DOC}+S_{AOD}+S_{BOC}\)
\(=2024^2+2025^2+2\cdot2024\cdot2025\)
\(=\left(2024+2025\right)^2=4049^2\left(đvdt\right)\)
