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Home » In the adjoining figure, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. Prove that (i) EF = FC (ii) AG : GD = 2 : 1
In the adjoining figure, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. Prove that
(i) EF = FC
(ii) AG : GD = 2 : 1
CBSE | Class 10 | maths | Maths | ML Aggarwal | Similarity
In the adjoining figure, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. Prove that
(i) EF = FC
(ii) AG : GD = 2 : 1

Solution:-

From the figure it is given that, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE.

(i) We have to prove that, EF = FC

From the figure, D is the midpoint of BC and also DF parallel to BE.

So, F is the midpoint of EC

Therefore, EF = FC

= ½ EC

EF = ½ AE

(ii) Now consider the ∆AGE and ∆ADF

Then, (BG or GE) ||DF

Therefore, ∆AGE ~ ∆ADF

So, AG/GD = AE/EF

AG/GD = 1/½

AG/GD = 1 × (2/1)

Therefore, AG: GD = 2: 1

i

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