Question

1. In the adjacent figure, ABC and DBC are two
triangles on the same side of BC. The mid-points of
AB, AC, DC, DB are E, E G and H respectively.
Prove that EFGH is a parallelogram.​

Answer 1

Answer:

IN TRIANGLE ABC

by mid point theorem

EF is // to BC and also EF=BC/2.................(1)

now in BCD

BY MID POINT THEOREM

HG// BC and also HG=BC/2....................(2)

from 1 and 2

EF//HG and EF=HG

HENCEEFGH is a parallelogram

join me on my youtube channel "A1 classes for mathematics"

for detailed explanation in videos according to your need

whatsapp7678664625

Answer 2

Answer:

apna desh mahan

subscribers ke liye log ajj kal kya kya karne lage hen