a) Ta có:
\(\left\{ \begin{array}{l}MB'\parallel (ABC )\\MB'\parallel (ABD)\\(ABC) \cap (ABD)=AB\end{array} \right. \)
\(\Rightarrow MB'\parallel AB\)
Do \(MB'\parallel AB\) nên \(MB’\) và \(AB\) xác định một mặt phẳng. Gọi \(MB\cup AB’\equiv I\).
Khi đó \(I \in BM \Rightarrow I \in \left( {BCD} \right)\)
\(I \in AB' \Rightarrow I \in \left( {ACD} \right)\)
Nên \(I \in \left( {BCD} \right) \cap \left( {ACD} \right) = CD\),
\(I \in CD\)
Vậy ba đường thẳng \(AB’\), \(BM\) và \(CD\) đồng quy tại \(I\).
b) \(MB'\parallel AB \Rightarrow \dfrac{MB'}{AB} = \dfrac{IM} {IB}\)
Kẻ \(MM' \bot CD\) và \(BH \bot CD\)
Ta có: \(MM'\parallel BH \Rightarrow \dfrac{IM}{IB} = \dfrac{MM'}{BH}\)
Mặt khác:
\(\left\{ \begin{array}{l}dt(\Delta MCD)=\dfrac{1}{2}CD.MM'\\dt(\Delta BCD)=\dfrac{1}{2}CD.BH\end{array} \right.\)
\(\Rightarrow\dfrac {dt(\Delta MCD)}{dt(\Delta BCD)}=\dfrac{\dfrac{1}{2}CD.MM'}{\dfrac{1}{2}CD.BH}\)
\(=\dfrac{MM'}{BH}\)
Do đó: \(\dfrac{MB'} {AB} = \dfrac{IM}{IB} \)
\(= \dfrac{MM'}{BH}= \dfrac {dt(\Delta MCD)}{dt(\Delta BCD)}\).
Vậy \(\dfrac{MB'}{AB} =\dfrac {dt(\Delta MCD)}{dt(\Delta BCD)}\).
c) Tương tự ta có:
\(\dfrac{MC'}{CA} =\dfrac {dt(\Delta MBD)}{dt(\Delta BCD)}\)
\(\dfrac{MD'}{DA} =\dfrac {dt(\Delta MBC)}{dt(\Delta BCD)}\)
Vậy:
\(\dfrac{MB'}{BA} + \dfrac{MC'}{CA} + \dfrac{M{\rm{D}}'}{DA}\)
\(=\dfrac {dt(\Delta MCD)}{dt(\Delta BCD)}+\dfrac {dt(\Delta MBD)}{dt(\Delta BCD)}+\dfrac {dt(\Delta MBC)}{dt(\Delta BCD)}\)
\(=\dfrac{dt(\Delta MCD)+dt(\Delta MBD)+dt(\Delta MBC)}{dt(\Delta BCD)}\)
\(=1\) .
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