Question

The line drawn through the midpoint of one side of a triangle parallel to another side bisect the third side​

Answer 1

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The line drawn through the midpoint of one side of a triangle parallel to another side bisect the third side

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${\blue{\sf\underline{Given}}}$

A ∆ABC in which D is the midpoint of AB and DE || BC.

${\blue{\sf\underline{To\:Prove}}}$

E is the midpoint of AC.

${\blue{\sf\underline{Construction}}}$

Draw CF || BA, meeting DE produced in F.

${\blue{\sf\underline{Propf}}}$

We have

⠀⠀⠀⠀DF || BC [∴DE || BC]

⠀⠀⠀⠀BD || CF [∴CF || BA]

∴ DBCF is a ||gm and so,

⠀⠀⠀⠀CF = DB = AD [∴D is the midpoint of AB].

Now,in ∆ADE and CFE, we have

⠀⠀⠀⠀∠EAD = ∠EFC (alt.interior ∠s)

⠀⠀⠀⠀AD = CF (proved)

and ∠ADE = ∠CFE (alt.interior ∠s)

∴ ∆ADE ≅ ∆CFE (ASA-criterion)

and so, AE = CE (c.p.c.t.).

Hence, E is the midpoint of AC.

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Answer 2

Given,In triangle ABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE||BC.

To prove, E is the midpoint of AC.

Since, D is the midpoint of AB

So,AD=DB

⇒ AD/DB=1.....................(i)

In triangle ABC,DE||BC,

By using basic proportionality theorem,

Therefore, AD/DB=AE/EC

From equation 1,we can write,

⇒ 1=AE/EC

So,AE=EC

Hence, proved,E is the midpoint of AC.