CAT Basic-Proportionality Theorem or Thales' Theorem MCQ

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Basic-Proportionality Theorem or Thales' Theorem

Theorem : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio.

Given: $\triangle A B C$ in which $D E \| B C$ and DE intersects AB and AC at D and E respectively.

To prove: $\frac{A D}{D B}=\frac{A E}{E C}$

Construction: Join BE and CD. Draw $E N \perp A B$ and $D M \perp A C$ .

Proof: We have

$\operatorname{ar}(\triangle A D E)=\frac{1}{2} \times A D \times E N \qquad\left[\because \Delta=\frac{1}{2} \times \text { base } \times \text { height }\right]$

$\text { and } \operatorname{ar}(\Delta D B E)=\frac{1}{2} \times D B \times E N$

$\therefore \quad \frac{\operatorname{ar}(\Delta A D E)}{\operatorname{ar}(\Delta D B E)}=\frac{\frac{1}{2} \times A D \times E N}{\frac{1}{2} \times D B \times E N}=\frac{A D}{D B}\qquad\qquad\ldots(i)$

$\text { Again, } \operatorname{ar}(\triangle A D E)=\operatorname{ar}(\triangle A E D)=\frac{1}{2} \times A E \times D M$

$\text { and } \operatorname{ar}(\Delta E C D)=\frac{1}{2} \times E C \times D M$

$\therefore \quad \frac{\operatorname{ar}(\Delta A D E)}{\operatorname{ar}(\Delta E C D)}=\frac{\frac{1}{2} \times A E \times D M}{\frac{1}{2} \times E C \times D M}=\frac{A E}{E C}\qquad\qquad\ldots(ii)$

Now, $\triangle D B E$ and $\triangle E C D$ being on the same base DE and between the same parallels DE and BC, we have

$\operatorname{ar}(\triangle D B E)=\operatorname{ar}(\triangle E C D)\qquad\qquad\ldots(iii)$

From (i), (ii) and (iii) we have

$\frac{A D}{D B}=\frac{A E}{E C}$

Theorem (Converse of Thales Theorem): If a line divides any two sides of a triangle in the same ratio then the line must be parallel to the third side.