In the adjoining figure, ABCD is a parallelogram. Line segments AX and CY bisect ∠A and ∠C respectively. Prove that :

(i) ΔADX ≅ ΔCBY

(ii) AX = CY

(iii) AX ∥ CY

(iv) AYCX is a parallelogram

Given,

ABCD is a parallelogram.

AX bisects ∠A

CY bisects ∠C

(i) In a parallelogram opposite angles are equal, thus ∠A = ∠C.

In ΔADX and ΔCBY,

∠D = ∠B [Opposite angles of a parallelogram are equal]

∠DAX = ∠BCY $\Big(\dfrac{1}{2}∠A = \dfrac{1}{2}∠C \Big)$

AD = BC [Opposite sides of a parallelogram are equal]

∴ ΔADX ≅ ΔCBY [By A.S.A. rule]

Hence, proved that ΔADX ≅ ΔCBY.

(ii) We know that,

ΔADX ≅ ΔCBY

AX = CY [By C.P.C.T.C.]

Hence, proved that AX = CY.

(iii) ∠DCY = ∠CYB [Alternate interior angles, AB ∥ DC, CY is transversal]

∠BAX = ∠DCY ($\dfrac{1}{2}∠A = \dfrac{1}{2}∠C$)

∴ ∠BAX = ∠CYB

These are corresponding angles formed by lines AX and CY with the transversal AB.

Since corresponding angles are equal, the lines AX and CY are parallel.

Hence, proved that AX ∥ CY.

(iv) We know that,

AX ∥ CY and AX = CY.

If one pair of opposite sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram.

Thus, AYCX is a parallelogram.

Hence, proved that AYCX is a parallelogram.