1.$\;\overline{MN}\;$ é paralelo a $\;\overline{AD}\;$ e $\;\overline{AD}\;$ é paralelo a $\;\overline{MP}\;$
$MN // AD\;\Rightarrow\;$ $\;\triangle BMN\thicksim\triangle BDA\;\Rightarrow\;\dfrac{MN}{DA}\,=\,\dfrac{BM}{BD}\;\Rightarrow\;$ $\;MN\,=\,DA\centerdot\, \dfrac{BM}{BD}\phantom{X}$(I)
$AD // MP\;\Rightarrow\;\triangle MPC\thicksim\triangle DAC\;\Rightarrow\;$ $\; \dfrac{MP}{DA}\,=\, \dfrac{MC}{DC}\;\Rightarrow\;$ $\;MP\,=\,DA\centerdot\,\dfrac{MC}{DC}\phantom{X}$(II)
2. Fazendo a soma (I) + (II):
$\;MN\,+\,MP\,=\,$ $\,DA\,\centerdot\,\dfrac{BM}{BD}\,+\,DA\,\centerdot\,\dfrac{MC}{DC}\;\Leftrightarrow\;$ $\;MN\,+\,MP\,=\,DA\,\centerdot\,(\dfrac{BM}{BD}\,+\, \dfrac{MC}{DC})$
3.$\;AD\;$ é a mediana relativa ao lado $\;BC\;\Rightarrow\;D\;$ é ponto médio de $\;BC\;\Rightarrow\;BD\,=\,DC\;$.
$\;MN\,+\,MP\,=\,DA\,\centerdot\,\left(\dfrac{BM}{BD}\,+\, \dfrac{MC}{BD}\right)\;\Leftrightarrow\;$ $\;MN\,+\,MP\,=\,DA\,\centerdot\,\left(\dfrac{BM + MC}{BD}\right)$
4. Da figura, $\;BM\,+\,MC\,=\,BC\;$, então concluimos que:
$\;MN\,+\,MP\,=\,DA\,\centerdot\,\left( \dfrac{BC}{BD}\right)\;\Leftrightarrow\;$ $\;MN\,+\,MP\,=\,DA\,\centerdot\, \dfrac{(BD\,+\,DC)}{BD}\;\Leftrightarrow$
$\Leftrightarrow\;MN\,+\,MP\,=\,DA\,\centerdot\,\dfrac{(BD\,+\,BD)}{BD}\;\Leftrightarrow\;$ $\;MN\,+\,MP\,=\,DA\,\centerdot\, \dfrac{2(BD)}{BD}\;\Leftrightarrow$$\Leftrightarrow\;MN\,+\,MP\,=\,DA\,\centerdot\,2\;\Leftrightarrow\;$
$\;\boxed{\;MN\,+\,MP\,=\,2\,\centerdot\,DA\;}$
