Question

State and prove midpoint theorem.

Answer 1

Hint: According to the midpoint theorem a line through the midpoint of one side of a triangle and parallel to another side, bisects the third side. Draw a line parallel to the side whose midpoint is taken from the point of intersection of the line and the third side. Use congruence rules to prove the theorem.

Complete step-by-step answer:

Statement of midpoint theorem: A line through the midpoint of one side of a triangle, parallel to another side bisects the third side.
Given: A triangle ABC. D is the midpoint of AC. DE||AB.
To prove: E is the midpoint of BC and 2DE = AB
Construction: Draw EF||AC

Proof:

In quadrilateral AFED we have
AD|| FE (By construction)
AF||DE (Given)
Hence AFED is a parallelogram.
Hence AD = EF ...(i)
But since D is the midpoint of AC we have AD = DC ...(ii)
From (i) and (ii) we get
EF = DC
Since AC || EF (by construction) we have
$\mathrm{\angle }DCE=\mathrm{\angle }FEB$ (corresponding angles)
Also, we have AD||EF
$\Rightarrow \mathrm{\angle }CAF=\mathrm{\angle }EFB$ (corresponding angles) ...(iii)
Since DE||AB, we have
$\mathrm{\angle }CAF=\mathrm{\angle }CDE$ (corresponding angles) ...(iv)
From equation (iii) and equation (iv), we get
$\mathrm{\angle }CDE=\mathrm{\angle }EFB$
Now In $\mathrm{\Delta }CDE$ and $\mathrm{\Delta }BEF$ we have
CD = FE (proved above)
$\mathrm{\angle }CDE=\mathrm{\angle }EFB$ (proved above)
$\mathrm{\angle }DCE=\mathrm{\angle }FEB$(proved above)
Hence $\mathrm{\Delta }CDE\cong \mathrm{\Delta }EFB$
Hence CE = EB (Corresponding parts of congruent triangles)
Hence E is the midpoint of BC
Also since $\mathrm{\Delta }CDE\cong \mathrm{\Delta }EFB$
DE = FB (Corresponding parts of congruent triangles) ...(v)
Since, AFED is a parallelogram.
DE = AF ...(vi)
Adding equation (v) and equation (vi), we get
2DE = AF+FB = AB
i.e. DE = AB
Hence proved.
Note: [1] If we join DF, then we divide the triangle into four smaller triangles. It can be shown that all four triangles are congruent to each other. Hence the area of triangle ABC = 4 times area of the triangle DEF
[2] The theorem can also be proved using vector algebra.
Let A (0),$$B\left(\overrightarrow{b}\right)$$ and $C\left(\overrightarrow{c}\right)$ be the position vector of each of the points A, B and C.
Now position vector of D $={\displaystyle \frac{0+\overrightarrow{c}}{2}}={\displaystyle \frac{\overrightarrow{c}}{2}}$
$\overrightarrow{AB}=P.V\left(B\right)-P.V\left(A\right)=\overrightarrow{b}-0=\overrightarrow{b}$
Hence the equation of DE is
$\overrightarrow{r}={\displaystyle \frac{\overrightarrow{c}}{2}}+t\overrightarrow{b}$
Equation of BC is $$\overrightarrow{r}=\overrightarrow{b}+u(\overrightarrow{b}-\overrightarrow{c})$$
E lies on both the lines DE and BC.
So, we have
$$\begin{array}{rl}& \overrightarrow{b}+u(\overrightarrow{b}-\overrightarrow{c})={\displaystyle \frac{\overrightarrow{c}}{2}}+t\overrightarrow{b}\\ & \Rightarrow (u+1-t)\overrightarrow{b}=({\displaystyle \frac{1}{2}}+u)\overrightarrow{c}\end{array}$$
Since $$\overrightarrow{b}$$ and $$\overrightarrow{c}$$ are Linearly independent, we have
$\begin{array}{rl}& u+{\displaystyle \frac{1}{2}}=0\\ & \Rightarrow u={\displaystyle \frac{-1}{2}}\end{array}$
and
$\begin{array}{rl}& u+1-t=0\\ & \Rightarrow t={\displaystyle \frac{1}{2}}\end{array}$
Hence the position vector of E is $${\displaystyle \frac{\overrightarrow{c}}{2}}+{\displaystyle \frac{1}{2}}\overrightarrow{b}={\displaystyle \frac{\overrightarrow{b}+\overrightarrow{c}}{2}}$$
which is the midpoint of BC.
Hence E is the midpoint of BC.
Now $$\left|\overrightarrow{DE}\right|=|{\displaystyle \frac{\overrightarrow{c}}{2}}-{\displaystyle \frac{\overrightarrow{b}+\overrightarrow{c}}{2}}|=\left|{\displaystyle \frac{\overrightarrow{b}}{2}}\right|={\displaystyle \frac{1}{2}}\left|\overrightarrow{b}\right|={\displaystyle \frac{1}{2}}\left|\overrightarrow{AB}\right|$$
i.e 2DE = AB
Hence proved.