Question

Question :-
State and prove mid point theorem
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kindly prove the mid point theorem using above diagram
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Answer 1

Step-by-step explanation:

it is in the attachment

hopes it will help✌️✌️

pls mark as brainliest !!!

Answer 2

$\small \rm \underline{The \: mid - point \: theorem \: is \: an \: useful \: fact} \\ \small \rm \underline{about \: the \: triangle. \: with \: the \: help \: of \: mid - } \\ \small \rm \underline{point \: theorem \: we \: can \: prove \: various \: results} \\ \small \rm \underline{related \: to \: quadrilateral \: and \: parallelogram}</p><p>$

Definition :-

$\rm \bold{Mid-point Theorem \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \rm \small {The \: line \: segment \: joining \: the \: mid - points} \\ \rm \small{of \: any \: two \: sides \: of \: a \: triangle \: is \: parallel \: to} \\ \rm \small{the \: third \: side \: and \: equal \: to \: half \: of \: it .\: \: \: \: \: \: \: \: \: \: \: }$

$\rm \bold{GIVEN \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \small \rm{In \: {\triangle ABC}, \: E \: and \: F \: are \: the \: mid \: points} \\ \rm \small{ of \: AB \: and \: AC, \: respectively \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

$\rm \bold{To \: Prove \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \small \rm{EF \parallel BC \: and \: EF = \frac{1}{2} \: BC}$

$\rm \bold{Construction \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ \rm \small{Through \: vertex \: C \: draw \: CD \parallel BA and \: let \: it} \\ \rm \small{meets \: extended \: EF \: at \: D \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

$\large \rm \underline{ \purple{proof}}$

$\rm{Now \: in \: \triangle AEF \: and \: \triangle CDF}$

$\rm{ \angle \: AFE = \angle \: CFD} \rm \tiny{  \left[ vertically \: opposite \: angles\right]}$

$\rm{AF = FC} \rm \tiny{  \left[ F, is \: the \: mid-point \: of \: AC \right]}$

$\rm{and \: \angle EAF = \angle \: FCD} \rm \tiny{  \left[ \because \: alternate \: interior \: angles \: \right]}$

$\rm{ \therefore \: \triangle \:AEF \cong \: \triangle CDF } \rm \small{  \left[ by \: ASA \: congruence \: rule\right]}$

$\rm \small{Then \: AE = CD \: and \: AE = FD} \\ </p><p> \rm \tiny{\left[ by \: CPCT\right]} \: \: \: \: \: \: \: \: \: \: eqation \: 1...$

$\rm{Also, \: AE = BE} \\ </p><p> \rm \tiny{\left[ E \: is \: the \: mid \: point \: of \: AB\right]} \: \: \: \: \: \: \: eqation \: 2...$

From equation (1) and (2) we get,

$\rm{AE = BE = CD} \: \: \: \: \: \rm \tiny{eqation \: 3....}$

$\rm{ and \: BE \parallel \: CD} \: \: \: \: \rm \tiny{(by \: construction)}$

so, BCDE is a parallelogram

$\rm \tiny{ \because \: one \: pair \: of \: opposite \: sides \: is \: parallel \: and \: equal}$

$\rm{Then, \: ED \parallel \: BC \: Or \: EF \parallel \: BC}$

$\rm{ and \: ED = BC}$

$\rm{Now, \: BC = ED \: = \: EF + FD} \rm \tiny{(from \: equation \: (1))}$

$\rm{ = 2 \: EF}$

$\small \rm \implies{EF \: = \frac{1}{2} \: BC}$

Hence,

the line segment joining the mid points of two sides of a triangle is parallel to the third side and equal to half of it.