मराठी

In the following figure, medians AD and BE of ΔABC meet at point G and DF || BE. Prove that: (i) EF = FC (ii) AG : GD = 2 : 1 - Mathematics

Advertisements
Advertisements

प्रश्न

In the following figure, medians AD and BE of ΔABC meet at point G and DF || BE. Prove that: EF = FC

बेरीज
Advertisements

उत्तर

Given:

In △ABC, AD and BE are medians, and DF || BE.

Since AD is the median of △ABC, D is the midpoint of BC.

Hence, BD = DC

Also, BE is the median of △ABC, so E is the midpoint of AC.

Therefore, points E, F, and C are collinear because F lies on AC.

Now consider △CBE,

In △CBE,

D is the midpoint of CB, and DF || BE.

By the Midpoint Theorem (the line through the midpoint of one side of a triangle and parallel to another side bisects the third side),

F is the midpoint of CE,

Therefore, CF = FE

That is,

EF = FC

Hence proved.

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 13: Similarity - Exercise 13A [पृष्ठ २७८]

APPEARS IN

नूतन Mathematics [English] Class 10 ICSE
पाठ 13 Similarity
Exercise 13A | Q 25. (i) | पृष्ठ २७८
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×