Question

prove that ABCD=1/2 EFGH​

Answer 1

Answer:

Given a parallelogram ABCD in which E,F,G,H rare mid points.

To prove: ABCD=1/2EFGH

Const. Join E AND G

Proof: In parallelogram EBCG triangle GEH is on same base and between same ll lines

so area GEH ,=1/2EBCG

similarly triangle GFE =1/2DAEG

add both

GEH+GFE= 1/2EBCG+1/2 DAEG

EFGH= 2 ABCD

1/2 EFGH= ABCD

proved