ABCD is a parallelogram and AX is parallel to CY. We need to prove: (i) $\angle AXB = \angle CYD$ (ii) $\triangle ABX \cong \triangle CYD$ (iii) AXCY is a parallelogram (iv) BY = XD

1. Problem Description

ABCD is a parallelogram and AX is parallel to CY. We need to prove:

(i) $\angle AXB = \angle CYD$

(ii) $\triangle ABX \cong \triangle CYD$

(iii) AXCY is a parallelogram

(iv) BY = XD

2. Solution Steps

(i) Prove that $\angle AXB = \angle CYD$

Since AX is parallel to CY and BC is a transversal, $\angle AXB = \angle XCY$ (corresponding angles).

In a parallelogram ABCD, AB is parallel to CD. So, BC is a transversal between the parallel lines AB and CD. Thus $\angle ABC = \angle BCD$.

Also, $\angle BCD = \angle YCD$

Since ABCD is a parallelogram, $\angle ABC = \angle ADC$. But $\angle CYD$ and $\angle ADC$ are supplementary.

Since ABCD is a parallelogram, AB || CD. So $\angle ABX = \angle CDY$ (alternate interior angles).

AX || CY. Thus, $\angle AXB$ and $\angle CYB$ are supplementary, as are $\angle CYD$ and $\angle AYD$.

Also, $\angle AXB$ and $\angle CYD$ are exterior angles.

Since AX is parallel to CY, $\angle AXB = \angle XCY$.

Since ABCD is a parallelogram, AB is parallel to CD.

So $\angle ABX = \angle YCD$ (alternate interior angles).

Also, $\angle B = \angle D$ as opposite angles of a parallelogram are equal.

Thus, $\angle ABX + \angle XBC = \angle YCD + \angle YDA$

Then, $\angle AXB = 180 - (\angle ABX + \angle BAX)$ and $\angle CYD = 180 - (\angle CDY + \angle DCY)$.

Since AB || CD, $\angle ABX = \angle CDY$.

Since AX || CY, $\angle BAX = \angle DCY$.

Therefore $\angle AXB = \angle CYD$.

(ii) Prove that $\triangle ABX \cong \triangle CYD$

Since ABCD is a parallelogram, AB = CD.

We have shown that $\angle AXB = \angle CYD$ in part (i).

Since ABCD is a parallelogram, $\angle ABX = \angle CDY$ (alternate interior angles)

Thus, by Angle-Angle-Side (AAS) congruence rule, $\triangle ABX \cong \triangle CYD$.

(iii) Prove that AXCY is a parallelogram

Since $\triangle ABX \cong \triangle CYD$, AX = CY.

We are given that AX || CY.

A quadrilateral with one pair of opposite sides equal and parallel is a parallelogram.

Therefore, AXCY is a parallelogram.

(iv) Show that BY = XD

Since $\triangle ABX \cong \triangle CYD$, BX = DY.

Since ABCD is a parallelogram, BC = AD.

BC = BX + XC and AD = AY + YD

Also, BC || AD.

Since ABCD is a parallelogram, BC = AD.

BC = BX + XC and AD = DY + YA.

Since $\triangle ABX \cong \triangle CYD$, BX = DY.

Thus BC - BX = AD - DY

XC = AY

Since AXCY is a parallelogram, XC || AY.

Since AXCY is a parallelogram, AC and XY bisect each other.

Also, BC = BX + XC, so BX = BC - XC

AD = DY + YA, so DY = AD - YA.

Since BC = AD and XC = YA, BX = DY.

BC = BY + YC, AD = AX + XD.

Since AXCY is a parallelogram, AX = CY.

So, AD - AX = BC - CY, which gives us XD = BY.

3. Final Answer

(i) $\angle AXB = \angle CYD$

(ii) $\triangle ABX \cong \triangle CYD$

(iii) AXCY is a parallelogram

(iv) BY = XD