In the adjoining figure, ABCD is a parallelogram and X, Y are the points on the diagonal BD such that DX = BY. Prove that
(i) CXAY is a parallelogram,
(ii) `triangle ADX ~= triangle CBY and triangle ABY ~= triangle CDX, and `
(iii) `AX = CY and CX = AY.`

Explanation

(i) Join AC , meeting BD at O.
Since the diagonals of a parallelogram bisect each other,
we have OA = OC and OD = OB.
Now, OD = OB and DX = BY
`rArr OD - DX = OB - BY rArr OX = OY.`
Now, OA = OC and OX = OY.
CXAY is a quadrilateral whose diagonals bisect each other.
`therefore CXAY is a || gm.`
(ii) and (iii) in `triangle ADX and CBY,` we have
`AD = BC " (opp. sides of a || gm)" `
`angle ADX = angle CBY " "("alt. interior " angle )`
`DX=BY " (given)" `
`therefore triangle ADX ~= triangle CBY " " `(SAS- criterion).
And so, `AX = CY " " `(c.p.c.t.).
Similarly, `triangle ABY ~= triangle CDX " and so " CX = AY `(c.p.c.t.).