Question extracted from RD sharma of class 10th, Chapter no. 4

Chapter name:- Triangles

Exercise :- 4.2

This is very basic and important questions. And has been asked in several times in exams.

In this question we have ΔABC,

In which D and E are points on the sides AB and AC respectively.

Now we have to show that DE ∥ BC

Also it is given AB= 12 cm, AD = 8 cm, AE = 12 cm, and AC = 18 cm.

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Given:Length of side AB = 12 cm, AD = 8 cm, AE = 12 cm, and AC = 18 cm

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

Length of side BD = AB – AD

⇒ 12 – 8 = 4 cm

And,

Length of side CE = AC – AE = 18 – 12 = 6 cm

Now,

AD/BD = 8/4 = 1/2 – equation 1

AE/CE = 12/6 = 1/2 – 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.