Question

please provide me proof of mid point theorem

Answer 1

given- a ∆ABC in which D and E are the mid points of sides AB and AC respectively. DE is joined.

to prove- DE || BC and DE = 1/2 BC

construction- produce the line segment DE to F such that DE=EF. join FC.

proof- in ∆s AED and CEF, we have
AE=CE (E is mid point of AC)
angle AED=CEF (vertically opposite angles)
DE=EF (by construction)
so ∆AED is congruent to ∆CEF (by SAS rule)
AD=CF (by cpct) ..........(i)
angle ADE=CEF (by cpct) ..........(ii)
now, D is the mid point of AB
so, AD=DB
and, DB=CF [from(i) AD=CF] ..........(iii)
now, DF intersects AD and FC at D and F respectively such that
angle ADE=CFE [from (ii)]
i.e., alternate interior angles are equal.
therefore, AD || FC
or, DB || FC ...........(iv)
from (iii) and (iv), we find that DBCF is a quadrilateral such that one pair of sides are equal and parallel.
therefore, DBCF is a parallelogram.
so, DF||BC and DF=BC [opposite side of ||gm are equal and parallel]

but D,E,F are collinear and DE=EF
therefore,DE||BC and DE=1/2BC

hence proved

Answer 2

Hi dear friend..
See the attached files


I hope it will help
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