Question

In the adjoining figure ,ABCD is a parallelogram ,E is the mid point of CD and through D,a line is drawn parallel to EB to meet CB produced at G and intersecting AB at F.

Prove that: (i) Ad=½GC (ii) DG=2EB

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Answer 1

Answer:

(i) In ∆ DCG, we have:

DG || EB

DE = EC (E is the midpoint of DC)

Also, GB = BC (By midpoint theorem)

∴ B is the midpoint of GC.

Now, GC = GB + BC

⇒ GC = 2BC

⇒ GC = 2 ⨯ AD [AD = BC]

∴ AD = 12GC

(ii) In ∆ DCG, DG || EB and E is the midpoint of DC and B is the midpoint of GC.

By midpoint theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and is half of it.

i.e., EB = 12DG

∴ DG = 2 ⨯ EB

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