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# Using Theorem 6.2, 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)

**Solution:**

We know that theorem 6.2 tells us if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. (Converse of Basic Proportionality theorem)

In ΔABC,

D is the midpoint of AB

⇒ AD = BD

AD/BD = 1............ (i)

E is the midpoint of AC

AE = CE

⇒ AE/CE = 1........ (ii)

From equations (i) and (ii)

AD/BD = AE/CE = 1

AD/BD = AE/CE

In ΔABC, according to theorem 6.2 (Converse of Basic Proportionality theorem),

Since, AD/BD = AE/CE

Thus, DE || BC

Hence, proved.

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 6

**Video Solution:**

## Using Theorem 6.2, 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 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 8

**Summary:**

Using Theorem 6.2, we have proved that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

**☛ Related Questions:**

- ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.
- The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.
- In Fig. 6.20, DE || OQ and DF || OR. Show that EF || QR.
- In Fig. 6.21, A, B and C are points on OP, OQ, and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

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