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Question
M is a point on the side BC of a parallelogram ABCD. DM when produced meets AB produced at N. Prove that
(1)` (DM)/(MN)=(DC)/(BN)`
(2)` (DN)/(DM)=(AN)/(DC)`
Solution
(i) Given: ABCD is a parallelogram
To prove :
(1)`(DM)/(MN)=(DC)/(BN)`
(2) `(DN)/(DM)=(AN)/(DC)`
Proof: In Δ DMC and Δ NMB
∠DMC = ∠NMB (Vertically opposite angle)
∠DCM = ∠NBM (Alternate angles)
By AAA- Similarity
ΔDMC ~ ΔNMB
∴`( DM)/(MN)=(DC)/(BM)`
NOW, `(MN)/(DM)+(BN)/(DC)`
Adding 1 to both sides, we get
`(MN)/(DM)+1=(BN)/(DC)+1`
⟹ `(MN+DM)/(DM)=(BN+DC)/(DC)`
⟹ `(MN+DM)/(DM)=(BN+AB)/(DC)` [∵ ABCD is a parallelogram]
⟹ `(DN)/(DM)=(AN)/(DC)`
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