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prove that the line segment joining the mid points of two sides of a triangle is parallel to the thurd side and half of it |
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Answer» 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 BCplease MARK it as brainliest. |
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