Giải thích các bước giải:
a.Ta có $AB//CD$
$\to S_{ACD}=S_{BCD}$
$\to S_{OAD}+S_{OCD}=S_{OBC}+S_{OCD}$
$\to S_{AOD}=S_{BOC}$
b.Ta có $AB//MN$
$\to \dfrac{OM}{AB}=\dfrac{DM}{DA}=\dfrac{CN}{CB}=\dfrac{ON}{AB}$
$\to OM=ON$
$\to O$ là trung điểm $MN$
$\to MN=2OM=2ON$
Ta có:
$\dfrac{OM}{AB}+\dfrac{ON}{CD}=\dfrac{DO}{DB}+\dfrac{OB}{BD}$
$\to \dfrac{\dfrac12MN}{AB}+\dfrac{\dfrac12MN}{CD}=\dfrac{DO+OB}{BD}$
$\to \dfrac{\dfrac12MN}{AB}+\dfrac{\dfrac12MN}{CD}=\dfrac{DB}{BD}$
$\to \dfrac{\dfrac12MN}{AB}+\dfrac{\dfrac12MN}{CD}=1$
$\to \dfrac{1}{AB}+\dfrac{1}{CD}=\dfrac{2}{MN}$
c.Ta có:
$\dfrac{S_{AOB}}{S_{AOD}}=\dfrac{OB}{OD}$
Vì $AB//CD\to \widehat{OAB}=\widehat{OCD},\widehat{OBA}=\widehat{ODC}$
$\to\Delta OAB\sim\Delta OCD(g.g)$
$\to \dfrac{S_{AOB}}{S_{COD}}=(\dfrac{OB}{OD})^2$
$\to \dfrac{S_{AOB}}{S_{COD}}=(\dfrac{S_{AOB}}{S_{AOD}})^2$
$\to \dfrac{S_{AOB}}{S_{COD}}=\dfrac{S^2_{AOB}}{S^2_{AOD}}$
$\to S^2_{AOD}=S_{COD}\cdot S_{AOB}$
Mà $S_{AOB}=2018^2, S_{COD}=2019^2$
$\to S_{AOD}=2018\cdot 2019$
$\to S_{OBC}=S_{AOD}=2018\cdot 2019$
$\to S_{ABCD}=S_{AOB}+2S_{AOD}+S_{OCD}$
$\to S_{ABCD}=2018^2+2\cdot 2018\cdot 2019+2019^2$
$\to S_{ABCD}=(2018+2019)^2$
$\to S_{ABCD}=4037^2$
