Question

. If one diagonal of a trapezium divides the other diagonal in the ratio 1 : 3. Prove that one of the parallel sides is three times the other.

Answer 1

Given that the diagonal BD the diagonal AC in AO : OC in 3 : 1
To prove : AB = 3CD
Proof : in Δ AOC and Δ DOC
Δ AOB = Δ COD
Δ OBA = Δ ODC (Because DC is parallel to AB, and DB is transversal so these are alternates)
Therefore,
Δ AOB is similar to COD (By AA similarity)
Now, AO/OC = AB/DC (Because in similar triangles sides are proportional)
3/1 = AB/DC (Given that AO : OC = 3 : 1)
So, AB = 3DC  
Hence proved.

Answer 2

Given : one diagonal of a trapezium divides other in tge ratio 1:3

Rtp: DC=3 AB

Proof: In triangle abo and triangle doc

Ang abo= ang doc(v.o.a)

Ang oba= ang odc( alternate angles)

Hence triangle abo is similar to triangle doc

By bpt bo/ od= ab/cd

Bo/od= 1/3 (according to question)

Hence. 1/3= ab/cd

Or 3ab= cd ( proved)