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In the given figure, ABCD is a quadrilat...

In the given figure, ABCD is a quadrilateral and E and F are points on AD and CD respectively such that AB = CB, `angleABE=angleCBF and angleEBD=angleFBD.` Prove that `BE=BF.`

Text Solution

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`angleABE=angleCBF and angleEBD=angleFBD`
`rArr" "angleABE+angleEBD=angleCBF+angleFBD`
`rArr" "angleABD+angleCBD." …(i)"`
Now, in `DeltaABD and DeltaCBD`, we have
`AB=CB" (given)"`
`angleABD=angleCBD" [from (i)]"`
`BD=BD" (common)"`
`therefore" "DeltaABD~=DeltaCBD" [SAS-criteria]."`
`therefore" "angleBAD=angleBCD" [c.p.c.t.]"`
`rArr" "angleBAE=angleBCF." ...(ii)"`
Now, in `DeltaABE and DeltaCBF`, we have
`AB=CB" (given)"`
`angleABE=angleCBF" (given)"`
`angleBAE=angleBCF" [proved in (ii)]"`
`therefore" "DeltaABE~=DeltaCBF" [AAS-criteria]."`
Hence, BE = BF.
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