Question

ABCD is a parallelogram.M is the midpoint of AC. X, Y are midpoints on AB and DC respectively such that AX=CY. Prove that
Triangle AXM is congruent to triangle CYM. And
XMY is a straight line.

Answer 1

Answer:

ΔAXM ≅ ΔCYM

XMY is straight line

Step-by-step explanation:

Let say XY cut AC at P

now Δ AXP  & ΔCYP

AX = CY   ( given)

AB ║ CD

=> ∠AXP = ∠CYM

∠XAP = ∠YCP

∠APX  = ∠CPY ( opposite Angle)

=> ΔAXP ≅ ΔCYP

=> AP = CP

=> AP + CP = AC

=> AP = CP = AC/2

=> P is mid pont of AC

given M is mid point of AC

=> P & M are same points

Hence

replacing P with M

ΔAXM ≅ ΔCYM

XPY is a straight line hence XMY is also straight line

Answer 2

Step-by-step explanation:

join XM and MY

️AXM and ️CYM

AM=MC ----(given)

AX =CY------(given)

angle XAM=Angle YCM (alternate angles)

therefore ️AXM=congruent ️CYM

2) angle AMX+angle AMY =180⁰ -------(linear pair of angle 180⁰)

therefore ,XMY is a straight line

hence proved