A(x1,y1),B(x2,y2) and C(x3,y3) are the vertices of triangle ABC and DEF are mid points
of the sides BC,CA,AB respectively.Prove that quadrilaetral DEFB is a parallelogram.

Dear Student,

Please find below the solution to the asked query:

We form our diagram from given information , As :


We know formula for coordinate of mid point ( x , y ) = $\left(\frac{x_1 + x_2}{2} , \frac{{\displaystyle y_1 + y_2}}{{\displaystyle 2}}\right)$

As D is mid point of BC , So , ( As we assume coordinate of D(x,y ) ), So

Coordinate of D =$\left(\frac{x_2 + x_3}{2} , \frac{{\displaystyle y_2+y_3}}{{\displaystyle 2}}\right)$

And

As E is mid point of CA , So , ( As we assume coordinate of E(x,y ) ), So

Coordinate of E =$\left(\frac{x_1 + x_3}{2} , \frac{{\displaystyle y_1+y_3}}{{\displaystyle 2}}\right)$

And

As F is mid point of AB , So , ( As we assume coordinate of F(x,y ) ), So

Coordinate of F =$\left(\frac{x_1 + x_2}{2} , \frac{{\displaystyle y_1+y_2}}{{\displaystyle 2}}\right)$


We know distance formula =  d  = $\sqrt{{\left(x_2 - x_1\right)}^2 + {\left(y_2 - y_1\right)}^2}$ , So

$$\begin{gathered} \mathrm{DE} = \sqrt{{\left(\frac{{\mathrm{x}}_1 + {\mathrm{x}}_3}{2} -\frac{{\mathrm{x}}_2 + {\mathrm{x}}_3}{2}\right)}^2 + {\left(\frac{{\mathrm{y}}_1 + {\mathrm{y}}_3}{2} -\frac{{\mathrm{y}}_2 + {\mathrm{y}}_3}{2}\right)}^2}= \sqrt{{\left(\frac{{\mathrm{x}}_1 - {\mathrm{x}}_2}{2}\right)}^2 + {\left(\frac{{\mathrm{y}}_1 - {\mathrm{y}}_2}{2} \right)}^2} =\mathbf{ }\frac{\sqrt{{\left({\mathit{x}}_{\mathbf{1}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathit{x}}_{\mathbf{2}}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\left({\mathit{y}}_{\mathbf{1}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathit{y}}_{\mathbf{2}}\right)}^{\mathbf{2}}}}{\mathbf{2}}\mathbf{ }\mathbf{unit} \\ \mathrm{EF} = \sqrt{{\left(\frac{x_1 + x_2}{2} -\frac{x_1 + x_3}{2}\right)}^2 + {\left(\frac{y_1 + y_2}{2} -\frac{y_1 + y_3}{2}\right)}^2}= \sqrt{{\left(\frac{x_2 - x_3}{2}\right)}^2 + {\left(\frac{y_2 - y_3}{2} \right)}^2} = \frac{\sqrt{{\left({\mathbf{x}}_{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathbf{x}}_{\mathbf{3}}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\left({\mathbf{y}}_{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathbf{y}}_{\mathbf{3}}\right)}^{\mathbf{2}}}}{\mathbf{2}}\mathbf{ }\mathbf{unit} \\ \mathrm{FB} = \sqrt{{\left(x_2 -\frac{x_1 + x_2}{2}\right)}^2 + {\left(y_2 -\frac{y_1 + y_2}{2}\right)}^2} =\sqrt{{\left(\frac{2 x_2 -x_1 - x_2}{2}\right)}^2 + {\left(\frac{2 y_2 -y_1 - y_2}{2}\right)}^2}= \sqrt{{\left(\frac{-\left(x_1 - x_2\right)}{2}\right)}^2 + {\left(\frac{-\left(y_1 - y_2\right)}{2} \right)}^2} =\mathbf{ }\frac{\sqrt{{\left({\mathbf{x}}_{\mathbf{1}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathbf{x}}_{\mathbf{2}}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\left({\mathbf{y}}_{\mathbf{1}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathbf{y}}_{\mathbf{2}}\right)}^{\mathbf{2}}}}{\mathbf{2}}\mathbf{ }\mathbf{unit} \\ \mathrm{BD} =\mathbf{ }\sqrt{{\left(\frac{x_2 + x_3}{2} -x_2\right)}^2 + {\left(\frac{y_2 + y_3}{2} -y_2\right)}^2}=\sqrt{{\left(\frac{x_2 +x_3 - 2x_2}{2}\right)}^2 + {\left(\frac{y_2 +y_3 - 2y_2}{2}\right)}^2}= \sqrt{{\left(\frac{-\left(x_2 - x_3\right)}{2}\right)}^2 + {\left(\frac{-\left(y_2 - y_3\right)}{2} \right)}^2}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\sqrt{{\left({\mathbf{x}}_{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathbf{x}}_{\mathbf{3}}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\left({\mathbf{y}}_{\mathbf{2}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathbf{y}}_{\mathbf{3}}\right)}^{\mathbf{2}}}}{\mathbf{2}}\mathbf{ }\mathbf{unit} \\ \mathrm{DF} =\mathbf{ }\sqrt{{\left(\frac{x_1 + x_2}{2} -\frac{x_2 + x_3}{2}\right)}^2 + {\left(\frac{y_1 + y_2}{2} -\frac{y_2 + y_3}{2}\right)}^2}= \sqrt{{\left(\frac{x_1 - x_3}{2}\right)}^2 + {\left(\frac{y_1 - y_3}{2} \right)}^2} =\mathbf{ }\frac{\sqrt{{\left({\mathbf{x}}_{\mathbf{1}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathbf{x}}_{\mathbf{3}}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{+}\mathbf{ }{\left({\mathbf{y}}_{\mathbf{1}}\mathbf{ }\mathbf{-}\mathbf{ }{\mathbf{y}}_{\mathbf{3}}\right)}^{\mathbf{2}}}}{\mathbf{2}}\mathbf{ }\mathbf{unit} \\ \mathrm{BE} =\sqrt{{\left(\frac{x_1 + x_3}{2} -x_2\right)}^2 + {\left(\frac{y_1 + y_3}{2} -y_2\right)}^2}=\sqrt{{\left(\frac{x_1 +x_3 - 2x_2}{2}\right)}^2 + {\left(\frac{y_1 +y_3 - 2y_2}{2}\right)}^2}\mathbf{ }\mathbf{=}\mathbf{ }\frac{\sqrt{{\left({\mathbf{x}}_{\mathbf{1}}\mathbf{ }\mathbf{+}{\mathbf{x}}_{\mathbf{3}}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{2}{\mathbf{x}}_{\mathbf{2}}\right)}^{\mathbf{2}}\mathbf{ }\mathbf{+}{\left({\mathbf{y}}_{\mathbf{1}}\mathbf{ }\mathbf{+}{\mathbf{y}}_{\mathbf{3}}\mathbf{ }\mathbf{-}\mathbf{ }\mathbf{2}{\mathbf{y}}_{\mathbf{2}}\right)}^{\mathbf{2}}}}{\mathbf{2}}\mathbf{ }\mathbf{unit} \end{gathered}$$

Here , We can see that opposite sides are equal to each other , As :

BD  =  EF    and DE =  FB                                                                            --- ( 1 )

Diagonals are  not equal to each other :  DF $\ne$BE                                    --- ( 2 )

From equation 1 and 2 we can say that :

DEFB is a parallelogram .                                                                               ( Hence proved )


Hope this information will clear your doubts about topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards