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    CAT 2026 Arithmetic Questions: Important Topics, Weightage, PYQs & Strategy
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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

    Quick Facts

    • 5 Questions around this concept.

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    QuestionEASY

     If three line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then all triangles formed is similar to each other. (True/False)

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    The shadow of a 5-m-long stick is 3 m long. At the same time the length of the shadow of a 12.5-m-high tree (in m) is:

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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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    Concepts Covered - 1

    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

    Proof: 

    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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