In the parallelogram ABCD, M is mid-point of AC and X, Y are points on AB and DC respectively such that AX = CY.

Prove that :

(i) triangle AXM is congruent to triangle CYM.

(ii) XMY is a straight line.

Given: In the parallelogram ABCD, M is mid-point of AC and X, Y are points on AB and DC respectively such that AX = CY.

To prove: (i) triangle AXM is congruent to triangle CYM.

(ii) XMY is a straight line.

Proof:

(i) In Δ AXM and Δ CYM,

AM = MC (given)

AX = CY (given)

∠XAM = ∠YCM (alternate angles, since AB || DC)

By SAS congruency criterion,

Δ AXM ≅ Δ CYM

Hence, triangle AXM is congruent to triangle CYM.

(ii) Since Δ AXM ≅ Δ CYM, we get:

∠AMX = ∠CMY

Since AM = MC, we also have:

∠AMX + ∠CMY = 180°

Thus, points X, M, Y are collinear, meaning:

XMY is a straight line.

Hence, XMY is a straight line.