[tex]\rm a)~ Fie~S-mijlocul~segmentului~[PN]. \\ \\ G_1-centru~de~greutate~in~ \Delta MNP \Rightarrow \frac{G_1S}{MS}= \frac{1}{3}.~~~~~~~~~~~~~~~~~(1) \\ \\ G_2-centru~de~greutate~in~\Delta QNP \Rightarrow \frac{G_2S}{QS} =\frac{1}{3}.~~~~~~~~~~~~~~~~~(2) \\ \\ Din~(1)~si~(2)~rezulta~G_1G_2~ ||~MQ~(teorema~reciproca~a~lui~Thales). \\ \\ b)~Din~punctul~anterior~rezulta~ca~\Delta MQS \sim \Delta G_1 G_2 S. ~(LUL) [/tex]
[tex]\rm Rezulta~ca ~ \frac{G_1 G_2}{MQ}= \frac{G_1 S}{MS}= \frac{1}{3} \Rightarrow G_1G_2 = \frac{MQ}{3}= \frac{18}{3}=6~(cm). [/tex]
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