Question

In the adjoining figure, ABCD is a parallelogram. E is the midpoint of DC and through D, a line segment is drawn parallel to EB to meet CB produced at G and it cuts AB at F. Prove that (i) $AD=\frac{1}{2}GC$, (ii) DG = 2EB.

Answer 1

(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 = $\frac{1}{2}GC$

(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 = ​$\frac{1}{2}DG$
∴​ DG = 2 ​⨯ EB