Question

Given triangle ABC with medians AE, BF, CD; FH parallel and equal in length to AE; BH and HE are drawn; FE extended to meet BH in G. Which one of the following statements is not necessarily correct?

(a) HE = HG 

(b) BH = DC 

(c) FG = \(\frac{3}{4}\) AB

(d) FG is the median of DBFH

Answer 1

(a) HE = HG

⇒ EH || AC, when extended meets AB in D. In congruent Δs ACD and HDB, DC = BH (corresponding sides) 

∴ (b) is true. 

Now FG = FE + EG

= \(\frac{1}{2}\)AB + \(\frac{1}{2}\) DB         (By mid-point theorem in Δs ABC and HDB respectively) 

= \(\frac{1}{2}\) AB + \(\frac{1}{2}\) x \(\frac{1}{2}\) AB       (∵ D is the mid-point of AB)

= \(\frac{3}{4}\) AB 

(c) is true. 

(d) FE || AB, when extended to G ⇒ EG || AB 

⇒ G is the mid-point of HB (By converse of mid-point theorem) 

∴ In ΔBFH, FG is the median, so (d) is also true.

(e) Cannot be proved true with the given information.