Category: Class 10 Maths most Important Topics

[mathjax] Basic Proportionality Theorem (Thales' Theorem) Statement : In a triangle, a line drawn parallel to one side of a triangle intersecting the other two sides in distinct points, divides the other two sides in the same ratio. Proof of the Theorem Given : $$\triangle A B C, $$ in which $$D E$$ is drawn parallel to $$B C$$ To Prove : $$\frac{A D}{D B}=\frac{A E}{E C}$$ Construction : Join $$C D$$ and $$B E$$ Draw $$D F \perp A E$$ and $$E G \perp A D$$ Proof $$ \begin{aligned} \operatorname{ar}(\triangle A D E) & =\frac{1}{2} \times A D \times E G \\ \operatorname{ar}(\triangle B D E) & =\frac{1}{2} \times B D \times E G \end{aligned} $$ Dividing (i) by (ii), we get $$ \frac{\operatorname{ar}(\triangle A D E)}{\operatorname{ar}(\triangle B D E)}=\frac{\frac{1}{2} \times A D \times E G}{\frac{1}{2} \times B D \times E G}=\frac{A D}{B D} $$ Similarly, $$ \begin{aligned} \operatorname{ar}(\triangle A D E) & =\frac{1}{2} \times D F \times A E \\ \operatorname{and} \operatorname{ar}(\triangle C D E) & =\frac{1}{2} \times C E \times D F \\ \Rightarrow \frac{\operatorname{ar}(\triangle A D E)}{\operatorname{ar}(\triangle C D E)}=\frac{\frac{1}{2} \times D F \times A E}{\frac{1}{2} \times D F \times C E} & =\frac{A E}{C E} \quad \ldots(\mathrm{iv} \end{aligned} $$ Now, $$\operatorname{ar}(\triangle B D E)=\operatorname{ar}(\triangle C D E)$$ $$\quad$$ Triangles on the same base and between the same parallel lines are equal in area $$ \Rightarrow \frac{\operatorname{ar}(\triangle A D E)}{\operatorname{ar}(\triangle B D E)}=\frac{\operatorname{ar}(\triangle A D E)}{\operatorname{ar}(\triangle C D E)} $$ $$\quad$$ From (iii) and (iv), we get $$ \frac{A D}{D B}=\frac{A E}{E C} $$ Hence proved.

Category: Class 10 Maths most Important Topics

Basic Proportionality Theorem (B.P.T)

Quick Links

Resources

Support