If one diagonal of the trapezium divides in the ratio 1:2

prove that one of the parallel side is double the other

Given : that diagonal BD divides diagonal AC in AO : OC with 3 : 1.

To prove : AB = 2 CD.

Proof : In triangle symbol AOC and triangle symbol DOC

 ∠AOB =    ∠COD

 ∠OBA =   ∠ODC ( because DC is parallel to AB and DB is the transversal, so these are alternate   ∠s )

 Therefore,  triangle symbol AOB   similar triangle symbol COD    [ By AA Similarity ]

Now, AO / OC = AB / DC                                ( because in similar triangles sides are propotional )

2 / 1  =  AB / DC                                               ( given that AO : OC = 2 : 1 )

AB = 2 DC

  • 15
Here is the answer to your question.
 
 
Given: ABCD is a trapezium. AB||CD. Diagonal BD divides the diagonal AC in the ratio 1:2 at O. i.e. OA: OC = 1:2.
 
To prove: CD = 2 AB
 
Proof: In ∆ AOB and ∆COD
 
∠AOB = ∠COD (Vertically opposite angles)
∠ABD = ∠ODC (Alternate angles)
 
Cheers!
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