Question

State and prove Basic proportionality (Thales) theorem.

Answer 1

Thales Theorem:

If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio

We consider a $\Delta ADE$ and a line segment BC which is parallel to DE.

In the figure above,

$area\left(\Delta ABC\right)=\frac{1}{2}\times AB\times CH\to \left(1\right)$

$area\left(\Delta ABC\right)=\frac{1}{2}\times AC\times BG\to \left(2\right)$

$area\left(\Delta ABE\right)=\frac{1}{2}\times AE\times BG\to \left(3\right)$

$area\left(\Delta ACD\right)=\frac{1}{2}\times AD\times CH\to \left(4\right)$

Also, $area\left(\Delta BCD\right)=area\left(\Delta BCE\right)\to \left(5\right)$

The above relation is true because both the triangles have a common base and are between the same 2 parallel lines.

Use relation$\left(5\right)$,

Add $area\left(\Delta ABC\right)$ to both sides of $\left(5\right)$

We get, $area\left(\Delta ACD\right)=area\left(\Delta ABE\right)\to \left(6\right)$

Divide $1$ by $4$

$=\frac{area\left(\Delta ABC\right)}{area\left(\Delta ACD\right)}=\frac{AB}{AD}$

Divide $2$ by $3$

$=\frac{area\left(\Delta ABC\right)}{area\left(\Delta ABE\right)}=\frac{AC}{AE}$

by using $\left(6\right)$,

$\frac{AB}{AD}=\frac{AC}{AE}$

Hence, Thales theorem is proved.