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सामन्तर चतुर्भुज ABCD में , BC और CD के मध्य - बिंदु क्रमश :E और F है , तो `vec (AE )` और `vec (AF )` को `vec (AB )` और `vec (AD )` के रूप में ज्ञात कीजिए और सिद्ध कीजिए कि -
`vec (AE )+vec (AF ) =(3 )/(2 ) vec (AC )`

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E ,BC का मध्य - बिंदु है ।
`vec (BE )= (1)/(2)vec (BC )=(1)/(2)vec(AD )`
image
`F,DC ` का मध्य - बिंदु ह|
`vec(DC ) =(1)/(2)vec(DC )=(1)/(2)vec(AB )`
`Delta ABE ` में ,
`vec(AE)=vec(AB)+vec(BE)=vec(AB)+(1)/(2)vec(AD)`
और `Delta ADF ` में ,
`vec(AF)=vec(AD)+vec(DF) =vec(AD )+(1)/(2)vec(AB)`
अब `vec(AE)+vec(AF)=vec(AB)+(1)/(2)vec(AD )+vec(AD)+(1)/(2)vec(AB)`
`=(3)/(2)(vec(AB)+vec(AD))=(3)/(2)(vec(AB)+vec(BC))`
`=(3)/(2) vec(AC)`

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