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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RD sharma, DHANPAT RAI publication
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.