Question

State and prove Thales Theorem.

Answer 1

Statement: If a line is drawn parallel to one side of a triangle, to interest the other two sides at distinct points, the other two sides are divided in the same ratio.

To prove: $\frac{AD}{BD}=\frac{AE}{CE}$

Consider $\Delta ABC$ .Let $DE\parallel BC$. Drop $FE$ and $DN$ perpendicular to sides $AB$ and $AC$ respectively.

Now,

Area of $\Delta ADE=\frac{1}{2}\times FE\times AD$ ...(i)

Area of $\Delta ADE=\frac{1}{2}\times AE\times DN$...(ii)

Also,

Area of $\Delta AEB=\frac{1}{2}\times FE\times AB$ ...(iii)

Area of $\Delta ADC=\frac{1}{2}\times AC\times DN$...(iv)

Now, since $\Delta BDE$ and $\Delta CED$ are on the same base $DE$ and between two parallel lines $DE$ and $BC$, therefore,

Area of $\Delta BDE=$ Area of $\Delta CED$

Adding area of $\Delta ADE$ on both the sides, we get,

Area of $\Delta BDE+\Delta ADE=$ Area of $\Delta CED+\Delta ADE$

$\Rightarrow$ Area of $\Delta AEB=$ Area of $\Delta ADC$...(v)

Now, (i) $÷$ (iii) , we get,

$\frac{ar\Delta ADE}{ar\Delta ADC}=\frac{\frac{1}{2}\times FE\times AD}{\frac{1}{2}\times FE\times AB}=\frac{AD}{AB}$ ...(vi)

Now, (ii) $÷$ (iv) , we get,

$\frac{ar\Delta ADE}{ar\Delta AEB}=\frac{\frac{1}{2}\times AE\times DN}{\frac{1}{2}\times AC\times DN}=\frac{AE}{AC}$ ...(vii)

From (v), (vi) and (vii), we get,

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

Subtracting $1$ from both sides, we get,

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

$\Rightarrow \frac{BD}{AD}=\frac{CE}{AE}$

Thus, $\frac{AD}{BD}=\frac{AE}{CE}$