Question

Question:....Trapezium ABCD with AB||DC.Point M is on the side AB and N on BC such that AM×NC=BN×MD.MN intersect AC at the point O. if AM/MD=⅔, find AO/OC.

=>Answer:​

Answer 1

Answer:

therefore the answer is AF/BF = ED/FC

Hence proved

Answer 2

Answer:

$Let a,b,c,d,m,n are the position vectors of A,B,C,D,M and N respectively</p><p></p><p>M and N are the midpoints of t diagonals AC and BD respectively ..... [Given]</p><p></p><p>∴m=2a+c and n=2b+d</p><p></p><p>⟹2m=a+c and 2n=b+d ........ (i)</p><p></p><p>Now, consider AB+AD+CB+CD</p><p></p><p>                         =(b−a)+(d−a)+(b−c)+(d−c)</p><p></p><p>                         =2b−2a−2c+2d</p><p></p><p>                         =2(b+d)−2(a+c)</p><p></p><p>                         =2(2n)−2(2m)</p><p></p><p>                         =4n−4m=4(n−m)=4MN</p><p></p><p>Hence, AB+AD+CB+CD=4MN</p><p></p><p></p><p>$