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In the figure given below, CD = ½ AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that : (i) CE || AG (ii) 3 ED = GD
In the figure given below, CD = ½ AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :
(i) CE || AG
(ii) 3 ED = GD
CBSE | Class 10 | maths | Maths | ML Aggarwal | Similarity
In the figure given below, CD = ½ AC, B is mid-point of AC and E is mid-point of DF. If BF || AG, prove that :
(i) CE || AG
(ii) 3 ED = GD

Solution:-

From the question it is given that,

CD = ½ AC

BF || AG

(i) We have to prove that, CE || AG 

Consider, CD = ½ AC

AC = 2BC … [because from the figure B is mid-point of AC]

So, CD = ½ (2BC)

CD = BC

Hence, CE || BF … [equation (i)]

Given, BF || AG … [equation (ii)]

By comparing the results of equation (i) and equation (ii) we get,

CE || AG

(ii) We have to prove that, 3 ED = GD

Consider the ∆AGD,

CE || AG … [above it is proved]

So, ED/GD = DC/AD

AD = AB + BC + DC

= DC + DC + DC

= 3DC

So, ED/GD = DC/(3DC)

ED/GD = 1/(3(1))

ED/GD = 1/3

3ED = GD

Hence it is proved that, 3ED = GD

i

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