Question

Theorem 89 : The line segment joining the mid-points of two sides of a triangle
is parallel to the third side.
You can prove this theo​

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: {\bigg \{ \fbox{mid \: point \: theorem}} \bigg \}$

${ \bold {✈︎{ \underline{ \underline{given}}}→ \: }}$

→ In a triangle ABC, D is the mid point of side AB and E is the mid point of side AC

__________________________

${ \bold {✈︎{ \underline{ \underline{to \: prove}}}→ \: }}$

→ DE parallel BC

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${ \bold {✈︎{ \underline{ \underline{construction}}}→ \: }}$

→ draw a line segment through C parallel to AB intersect DE produced at F

__________________________

${ \bold {✈︎{ \underline{ \underline{proof}}}→ \: }}$

→ in ∆ADE and ∆CEF

ᗒ angle AED = angle CEF (vertical opposite angles)

ᗒ AE = CE ( since E is the mid point of AC)

ᗒ angle DAE = angle FCE ( alternate angles)

$\bold{ \tt \implies \:∆ADE \cong ∆CEF }$

( by ASA congruence rule)

→ AD = CF (by CPCT) ............(i)

→ DE = EF (by CPCT) ............(ii)

{ CPCT = congruent parts of congruent triangles}

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we have,

AD = BD ( since D is the mid point of AB)

and from (i) we have,

AD = CF

→ BD = CF ..............(iii)

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AB || CF (by construction)

→ CF || BD ( since BDA || CF ) ............(iv)

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now, in quadrilateral BCFD ;

ᗒ BD = CF [from (iii)]

ᗒ BD || CF [from (iv)]

→ BCFD is a parallelogram ( theorem: since a pair of opposite sides of a quadrilateral are parallel and equal to each other then, that quadrilateral is a parallelogram)

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since, BCFD is a parallelogram

so that, DF || BC and DF = BC ( opposite sides of parallelogram are parallel and equal to each other )

since, DF || BC

→ DE || BC

and DF = BC

→ DE + EF = BC

→ 2DE = BC [ since DE = EF from (ii) ]

→ DE = BC/2

➜ hence, proved that the line segment joining the mid points of two sides of a triangle is parallel to the third side and half of it