Question

Question 8 Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Class 10 - Math - Triangles Page 129

Answer 1

Basic proportionality theorem

If a line is drawn parallel to one side of a triangle to intersect the other two side interesting points than the other two sides are divided in the same ratio and  this theorem is also known as Thales theorem.

Converse of basic proportionality theorem

If a line divides any two sides of a triangle in the same ratio then the line must parallel to the third side.

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Given:

ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.

To Prove: DE || BC

Proof: D is the mid point of AB (Given)

∴ AD=DB

⇒ AD/BD = 1 … (i)

Also, E is the mid-point of AC (Given)
∴ AE=EC

⇒AE/EC = 1 [From equation (i)]

From equation (i) and (ii), we get

AD/BD = AE/EC

∴ DE || BC [By converse of Basic Proportionality Theorem]

Hence , the line joining the mid points of any two sides of a triangle is parallel to the third side....

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Hope this will help you....

Answer 2

Given: ΔABC in which D and E are the mid points of AB and AC respectively such that AD=BD and AE=EC.

To Prove: DE || BC

Proof: D is the mid point of AB (Given)

∴ AD=DB

⇒ AD/BD = 1 ... (i)


Also, E is the mid-point of AC (Given)
∴ AE=EC

⇒AE/EC = 1 [From equation (i)]

From equation (i) and (ii), we get
AD/BD = AE/EC
Hence, DE || BC [By converse of Basic Proportionality Theorem]