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

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