Question

M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that
` (i) (DM)/(MN)=(DC)/(BN)" " (ii) (DN)/(DM)=(AN)/(DC)`

Answer 1

In `Delta NDA, MB||DA`,
`rArr (NM)/(DM)=(NB)/(BA)rArr (DM)/(MN)=(AB)/(BN)rArr (DM)/(MN)=(DC)/(BN) [ :. AB=DC]`
(ii) `rArr (NM)/(DM)=(NB)/(BA)rArr (NM)/(MD)=+1=(NB)/(BA)+1rArr (NM+MD)/(MD)=(NB+BA)/(BA)`
`(DN)/(DM)=(AN)/(AB)rArr (DN)/(DM)=(AN)/(DC) [ :. AB=DC]`