Cho hình thang ABCD (AB//CD), gọi O là giao điểm 2 đường chéo. CMR: Diện tích OAB + Diện tích OCD ≥ 1/2 Diện tích ABCD

Giải thích các bước giải:

Qua O kẻ  \(OH \bot AB\,\,\left( {H \in AB} \right);\,\,\,OK \bot CD\,\,\left( {K \in CD} \right)\). Khi đó ta có:

\(\begin{array}{l}
{S_{OAB}} + {S_{OCD}} = \frac{1}{2}OH.AB + \frac{1}{2}OK.CD = \frac{1}{2}\left( {OH.AB + OK.CD} \right)\\
{S_{ABCD}} = \frac{1}{2}HK.\left( {AB + CD} \right) = \frac{1}{2}\left( {OH + OK} \right)\left( {AB + CD} \right)\\
\frac{1}{2}{S_{ABCD}} – \left( {{S_{OAB}} + {S_{OCD}}} \right)\\
 = \frac{1}{4}.\left[ {\left( {OH + OK} \right).\left( {AB + CD} \right) – 2\left( {OH.AB + OK.CD} \right)} \right]\\
 = \frac{1}{4}\left( {OH.AB + OH.CD + OK.AB + OK.CD – 2\left( {OH.AB + OK.CD} \right)} \right)\\
 = \frac{1}{4}\left( {OH.CD + OK.AB – OH.AB – OK.CD} \right)\\
 = \frac{1}{4}\left[ {OH\left( {CD – AB} \right) + OK\left( {AB – CD} \right)} \right]\\
 = \frac{1}{4}\left( {OH – OK} \right)\left( {CD – AB} \right)\\
AB//CD \Rightarrow \frac{{OH}}{{OK}} = \frac{{OA}}{{OC}} = \frac{{AB}}{{CD}}\\
AB < CD \Rightarrow OH < OK \Rightarrow \left( {OH – OK} \right)\left( {CD – AB} \right) < 0\\
AB > CD \Rightarrow OH > OK \Rightarrow \left( {OH – OK} \right)\left( {CD – AB} \right) < 0\\
 \Rightarrow \frac{1}{2}{S_{ABCD}} – \left( {{S_{OAB}} + {S_{OCD}}} \right) < 0\\
 \Rightarrow \frac{1}{2}{S_{ABCD}} < {S_{OAB}} + {S_{OCD}}
\end{array}\)