Question

The line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it. -Maths 9th

Answer 1

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