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

(i) Given: ABCD is a parallelogram 

To prove: 

(i) `(DM)/(MN) = (DC)/(BN)` 

(ii) `(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)`