Prove analytically that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.


To do:

We have to prove that the line segment joining the middle points of two sides of a triangle is equal to half of the third side.

Solution:

Let $A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)$ be the vertices of a $\triangle ABC$.
Let $D$ and $E$ be the mid-points of the sides $AB$ and $AC$ respectively.

This implies,

\( \mathrm{DE}=\frac{1}{2} \mathrm{BC} \)

The coordinates of \( \mathrm{D} \) are \( \left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right) \)

The coordinates of \( \mathrm{E} \) are \( \left(\frac{x_{1}+x_{3}}{2}, \frac{y_{1}+y_{3}}{2}\right) \)

The length of the side $BC=\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$......(i)

The length of the side $DE=\sqrt{\left(\frac{x_{1} +x_{3}}{2} -\frac{x_{1} +x_{2}}{2}\right)^{2} +\left(\frac{y_{1} +y_{3}}{2} -\frac{y_{1} +y_{3}}{2}\right)^{2}}$

$=\sqrt{\frac{(x_1+x_3-x_1-x_2)^2}{4}+\frac{(y_1+y_3-y_1-y_2)^2}{4}}$

$=\sqrt{\frac{( x_{3} -x_{2})}{4}^{2} +\frac{( y_{3} -y_{2})}{4}^{2}}$

$=\frac{1}{2}\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$

$=\frac{1}{2}BC$    (From (i))

Hence proved. 

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Updated on: 10-Oct-2022

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