Question

If one diagonal of a trapizum divides the other diagonal in the ratio 1:3. Prove that one of the parallel side is 3 times the other

Answer 1

Answer:

proved below

Step-by-step explanation:

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

Hello mate!!

.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)


Hope it helpful