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A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
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
DE || AC
BD/AD = BE/EC .........(i)
In ΔABE
DF || AE
BD/AD = BF/FE ........(ii)
From (i) and (ii)
BD/AD = BE/EC = BF/FE
Thus, BE/EC = BF/FE
☛ Check: NCERT Solutions Class 10 Maths Chapter 6
Video Solution:
In Fig. 6.19, DE || AC and DF || AE. Prove that BF/FE = BE/EC.
Class 10 Maths NCERT Solutions Chapter 6 Exercise 6.2 Question 4
Summary:
In the figure, if DE || AC and DF || AE we have proved that BF/FE = BE/EC.
☛ Related Questions:
- 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).
- 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).