Question

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.​

Answer 1

AnswEr :

⋆ Refrence of Image is in the Diagram :

$\setlength{\unitlength}{2cm}\begin{picture}(16,4)\thicklines\put(8.8,3){A}\put(7.8,1){B}\put(10.1,1){C}\put(8,1){\line(1,0){2}}\put(8,1){\line(1,2){1}}\put(10,1){\line(-1,2){1}}\put(10,1){\line(1,2){0.5}}\put(8.5,2){\line(1,0){2}}\put(8.2,2){D}\put(9.56,2.04){E}\put(10.56,2){F}\qbezier(9.85,1.3)(9.97,1.4)(10.12,1.28)\qbezier(9.22,2)(9.1,2.19)(9.4,2.2)\qbezier(9.6,1.8)(10,2)(9.8,2)\end{picture}$

$\rule{170}{1}$

$\bf{\dag}\:\underline{\frak{Given :}}\\\\\textsf{Consider the triangle ABC, as shown in Diagram :}\\\\\textsf{Let E and D be the midpoints of the sides AC and AB.}\\\textsf{Then the line DE is said to be parallel to the side BC,}\\\textsf{whereas the side DE is half of the side BC i.e.}\\\\\\\bf{\dag}\:\underline{\frak{To\:Prove :}}\\\\\sf1)\:DE\parallel BC\\\sf2)\:DE=\dfrac{1}{2}\times BC\\\\\\\bf{\dag}\:\underline{\frak{Construction :}}\\\\\textsf{Extend the line segment DE and produce it to}\\\textsf{F such that, EF = DE}$

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$\bigstar\underline{\textsf{ In \triangle ADE and, \triangle CFE:}}\\\twoheadrightarrow\:\tt EC = AE\qquad\:\:\:\qquad(Given)\\\twoheadrightarrow\:\tt\angle CEF=\angle AED\:\qquad(Vertically\:Opp.\:\angle s)\\\twoheadrightarrow\:\tt EF=DE\qquad\qquad\:\:\:\:\!(Construction)\\\\\therefore\: \boxed{\tt {\scriptsize\triangle}\:CFE \cong{\scriptsize\triangle}\:ADE}\\ \dfrac{\qquad\qquad \qquad\qquad\qquad}{} \\\bigstar\:\underline{\sf BY\:\:CPCT :}\\\tt1)\:\angle CFE=\angle ADE\\\tt2)\:\angle FCE=\angle DAE\\ \tt3)\:CF=AD$

$\textsf{The angles, \angle CFE and \angle ADE are the alternate}\\\textsf{interior angles. Assume CF and AB as two lines}\\\textsf{which are intersected by the transversal DF.}\\\\\textsf{In a similar way, \angle FCE and \angle DAE are the alternate}\\\textsf{interior angles. Assume CF and AB are the two lines}\\\textsf{which are intersected by the transversal AC.}$

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$\begin{aligned}\sf Therefore\quad CF \parallel AB \\\sf So\quad CF \parallel BD\\\sf\&\quad CF = BD \end{aligned}\\\textsf{\lbrack since BD = AD, it is proved that CF = AD\rbrack}\\\\\textsf{Thus, BDFC forms a parallelogram.}\\\\\star\:\texttt{By the use of properties of a}\\\texttt{parallelogram, we can write :} \\\\:\implies\sf BC\parallel DF\quad\&\quad BC = DF$

$\boxed{\begin{minipage}{6 cm}\bf{\dag}\:\underline{we will get :}\\\sf\\1) DE \parallel BC\\\\2) DE = \dfrac{1}{2}\times BC\\\\\\\underline{Hence, Midpoint Theorem is Proved.}\end{minipage}}$

Answer 2

Answer:

Proof: Consider triangle ABC.

Join mid-points of AB and AC.

Let mid-point of AB be M and mid-point of BC be N.

Now we have two triangles ABC and AMN

In triangle ABC and triangle AMN

AB/AM=AC/AN=2..........by construction

angle BAC=angle MAN.........as A-M-B and A-N-C.

thus triangle ABC∼∼ triangle AMN.

Thus angle AMN=angle ABC as triangles are similar

and MN ∥∥ BC as corresponding angles are equal.

Also MN/BC = AM/AB = 1/2 (sides of similar triangle.)