IF ONE DIAGONAL OF A TRAPEZIUM DIVIDES THE OTHER DIAGONAL IN THE RATIO 1:2.PROVE THAT ONE OF THE PARALLEL SIDES IS DOUBLE THE OTHER
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Given, that the diagonal BD divides the AC in AO : OC with 2 : 1
To prove : AB = 2C
Proof : In
Δ AOC and Δ DOC
AOB = COD
OBA = ODC (Because DC is parallel to AB, and DB is transversal. So, these are alternates)
Therefore,
Δ AOB is same as Δ COD (By AA similarity)
Now, AO/OC = AB/DC (Because in similar triangles sides are proportional)
⇒ 2/1 = AB/DC (Given that AO : OC = 2 : 1)
So, AB = 2CD
Hence proved.
Given, that the diagonal BD divides the AC in AO : OC with 2 : 1
To prove : AB = 2C
Proof : In
Δ AOC and Δ DOC
AOB = COD
OBA = ODC (Because DC is parallel to AB, and DB is transversal. So, these are alternates)
Therefore,
Δ AOB is same as Δ COD (By AA similarity)
Now, AO/OC = AB/DC (Because in similar triangles sides are proportional)
⇒ 2/1 = AB/DC (Given that AO : OC = 2 : 1)
So, AB = 2CD
Hence proved.
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