Question

ABCD patrulater M inclus in [BC] , N inclus in [AC] , P inclus in [CD] iar NM paralel AB , NP paralel AD

MP // BD?
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Answer 1

$\it Din\ \Delta ABC \Rightarrow MN||AB\ \stackrel{T.Thales}{\Longrightarrow}\ \dfrac{CM}{MB}=\dfrac{CN}{NA}\ \ \ \ \ (1)\\ \\ \\ Din\ \Delta ACD \Rightarrow NP||AD\ \stackrel{T.Thales}{\Longrightarrow}\ \dfrac{CN}{NA}=\dfrac{CP}{PD}\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow \dfrac{CM}{MB}=\dfrac{CP}{PD}\ \ \ \ \ (3)$

Din  relația  (3), cu reciproca teoremei lui Thales în ΔBCD, ⇒MP || BD .