A line segment (DE) joining the midpoints of two sides of a triangle is parallel
to the third sid

Given = A △ABC in which D and E are the mid-points of side AB and AC respectively. DE is joined .

To Prove : DE || BC and DE = 1 / 2 BC.

Const. : 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 the mid point of AC]

∠AED = ∠CEF[vert. opp.∠s]

and DE = EF [by construction]

∴ △AED ≅ △CEF [by SAS congruence axiom]

⇒ AD = CF ---(i)[c.p.c.t.]

and ∠ADE and ∠CEF ---(ii) [c.p.c.t.]

Now, D is the mid point of AB.

⇒ AD = DB ---(iii)

From (i) and (iii), CF = DB ---(iv)

Also, from (ii)

⇒ AD = || FC [if a pair of alt. int. ∠s are equal then lines are parallel]

⇒ DB || BC ---(v)

From (iv) and (v), we find that DBCF is a quadrilateral such that one pair of opposite sides are equal and parallel.

∴ DBCF is a ||gm

⇒ DF || BC and DF = BC [∵opp side of ||gm are equal and parallel]

Also, DE = EF [by construction]

Hence, DE || BC and DE = 1 / 2 BC

step-by-step explanation:

answers available in this download

(pearson%2c%25202012)%2520bbs.pdf& usg=aovvaw1obmtxnfghfgxhx0urh_je

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