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# In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.

**Solution:**

We know according to the basic proportionality theorem, if a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, the other two sides are divided in the same ratio.

In ΔABC

LM || CB

AM/MB = AL/LC............ (1)

In ΔACD

LN || CD

AN/DN = AL/LC............ (2)

From equations (1) and (2)

AM/MB = AN/DN

⇒ MB/AM = DN/AN

Adding 1 on both sides

MB/AM + 1 = DN/AN + 1

(MB + AM)/AM = (DN + AN)/AN

AB/AM = AD/AN

⇒ AM/AB = AN/AD

Hence proved.

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

**Video Solution:**

## In Fig. 6.18, if LM || CB and LN || CD, prove that AM/AB = AN/AD.

Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 3

**Summary:**

If LM || CB and LN || CD, it is proved that AM/AB = AN/AD

**☛ Related Questions:**

- In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
- 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.
- Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

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