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In a ΔABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE ∥ BC:AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.

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Question taken from RD sharma
Class 10th
Chapter no. 4
Chapter name:- Triangles
Exercise :- 4.2
This is very basic and important questions.

We have been given that in ΔABC, D and E are points on the sides AB and AC respectively

Now For the following case we have to  show that DE ∥ BC

It is given that AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.

 

Understanding and learning CBSE maths
RD sharma, DHANPAT RAI publication

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1 Answer

  1. Given:

    Length of side AB = 10.8 cm, BD = 4.5 cm, AC = 4.8 cm, and AE = 2.8 cm.

    To show: Side DE is parallel to BC ( DE || BC )

    Now,

    Length of side AD = AB – DB = 10.8 – 4.5 = 6.3

    And,

    Length of side CE = AC – AE = 4.8 – 2.8 = 2

    Now,

    AD/BD = 6.3/ 4.5 = 2.8/ 2.0 = 7/5                   – equation 1

    AE/CE = 7/5                                                   – equation 2

    Now from equation 1 and 2

    AD/BD = AE/CE

    So, by the converse of Thale’s Theorem

    We get, DE ∥ BC.

    Hence Proved

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