Question

L and M are the mid points of sides AB and DC respectively of parralleologram ABCD. Prove that segments DL and BM trisects diagonal AC. ​

Answer 1

From figure,

BL = DM and BL || DM and BLMD is a

parallelogram, therefore BM || DL

From triangle ABY

L is the midpoint of AB and XL || BY,

therefore x is the midpoint of AY

.ie AX = XY .....$(1)$

Similarly for triangle CDX

CY=XY .....$(2)$

From $(1)$ and $(2)$

AX = XY = CY and AC = AX + XY + CY

Hence proved.

Answer 2

Answer:

Step-by-step explanation:

BL=DM and BL II DL and BLMD is a parallelogram ,therefore BM II DL

from triangle ABY

L is the midpoint of AB and XL II BY

x is the midpoint of AY ie. AX =XY .....(1)

similarly for triangle CDX

CY = XY ...(2)

from (1) and (2)

AX = XY= CY and AC = AX + XY + CY

Hence proved