ABCD is a //gm and line segment AX and CY bisects angles A and C respectively where X is a point on AB. To prove AX // CY
Given, ABCD is a parallelogram.
∴ ∠A = ∠C (Opposite angles of the parallelogram are equal)
⇒ ∠1 = ∠2 ...(1) [AX is bisector of ∠A and CY is bisector of ∠C]
Now, AB || CD and CY is the transversal.
∴ ∠2 = ∠3 ...(2) [Alternate interior angles]
From (1) and (2), we get
∠1 = ∠3
∴ Transversal AB intersects AX and CY at A and Y respectively such that ∠1 = ∠3 i.e., corresponding angles formed are equal.
∴ AX || CY.