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CAT Basic-Proportionality Theorem or Thales' Theorem - Practice Questions & MCQ

Edited By admin | Updated on Oct 04, 2023 04:20 PM | #CAT

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ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB then 

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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.

Given: \triangle A B C and a line l intersecting AB at D and AC at E, such that \frac{A D}{D B}=\frac{A E}{E C}.

To Prove: D E \| B C


If possible, let DE not be parallel to BC. Then, there must be another line through D which is parallel to BC.

\text{Let }D F \| B C

Then, by Thales' theorem, we have

\frac{A D}{D B}=\frac{A F}{F C}

\text{But, }\frac{A D}{D B}=\frac{A E}{E C}\qquad(\text { given })

From (i) and (ii), we get

        \frac{A F}{F C}=\frac{A E}{E C}

\\ \Rightarrow \frac{A F}{F C}+1=\frac{A E}{E C}+1

\Rightarrow \frac{A F+F C}{F C}=\frac{A E+E C}{E C} \\

\Rightarrow \frac{A C}{F C}=\frac{A C}{E C} \Rightarrow \frac{1}{F C}=\frac{1}{E C} \Rightarrow F C=E C

This is possible only when E and F coincide.

Hence, D E \| B C

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