Question

Prove that the line segment joining the mid-points of two sides of a

triangle is parallel to the third side and half of it.​

Answer 1

Answer:

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Step-by-step explanation:

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

Answer 2

Answer:

The midpoint theorem states that “The line segment in a triangle joining the midpoint of two sides of the triangle is said to be parallel to its third side and is also half of the length of the third side

Step-by-step explanation:

If the line segment adjoins midpoints of any of the sides of a triangle, then the line segment is said to be parallel to all the remaining sides, and it measures about half of the remaining sides.

Consider the triangle ABC, as shown in the above figure,

Let E and D be the midpoints of the sides AC and AB. Then the line DE is said to be parallel to the side BC, whereas the side DE is half of the side BC; i.e.

DE∥BC

DE = (1/2 *  BC).