Question

The line segment joining the midpoint of two sides of a triangle is parallel to the third side and equal to half of it. Prove this theorem​

Answer 1

$\bf\underline{\underline{\pink{Question:-}}}$

★ The line segment joining the midpoint of two sides of a triangle is parallel to the third side and equal to half of it. Prove this theorem.

$\bf\underline{\underline{\blue{Given:-}}}$

★ A △ABC in which D and E are the midpoints of AB and AC respectively. DE is joined.

$\bf\underline{\underline{\green{To Prove:-}}}$

★ DE || BC and DE = ½BC

$\bf\underline{\underline{\red{Construction:-}}}$

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

$\bf\underline{\underline{\orange{Proof:-}}}$

In △AED and CEF, we have :

∠AED = CEF ( vert. opp. angle)

AE = CE [∵ E is the midpoint of AC]

∠DAE = ∠FCE (alt. interior angle)

∴△AED ≅△FCE (ASA-criterion)

And so, AD = CF and DE = EF (c.p.c.t).

But, AD = BD [∵ D is the midpoint of AB]

and BD || CF (by construction)

∴ BD = CF and BD || CF

==> BCDF is a ||gm

==> DF || BC and DF = BC

==> DE || BC and DE = ½DF = ½BC [∵ DE = EF].

Hence, DE || BC and DE = ½BC

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Answer 2

Given :  The line segment joining the midpoint of two sides of a triangle is parallel to the third side and equal to half of it.

To Find : Prove this theorem​

Solution:

Consider  ΔABC

D is id point of AB and E is mid point of AC

To be Proved

DE = BC/2  and DE || BC

in Δ ABC  and  Δ ADE

AD/AB  = AE/AC  = 1/2    as  D and E are mid points of AB and AC respectively

∠A = ∠A   common angle

Hence Using SAS Similarity

Δ ABC  ~  Δ ADE

SAS Similarity

If an angle of one triangle and an angle of another triangle have the same measure, and the sides containing those angles are proportional, then the triangles are similar.

Δ ABC  ~  Δ ADE

AD/AB  = AE/AC = DE/BC

=> DE/BC = 1/2

=> DE = BC/2

QED

Hence proved

Δ ABC  ~  Δ ADE

=> ∠B = ∠D  and  ∠C = ∠E

Corresponding angles are equal

Two lines that form congruent corresponding angles with a transversal are parallel.

Hence DE || BC

QED

Hence Proved

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