Question

In parallelogram ABCD, E and F are mid-points of the sides AB and CD respectively. The line segments AF and BF meet the line segments ED and EC at points G and H respectively.
Prove that:
(i) Triangles HEB and FHC are congruent;
(ii) GEHF is a parallelogram.

Answer 1

The figure is shown below

(i) From ΔHEB and ΔFHC
BE = FC
∠EHB = ∠FHC                        ...[ Opposite angle ]
∠HBE = ∠HFC
∴  ΔHEB ≅ ΔFHC
∴  EH = CH , BH = FH

(ii) Similarly AG = GF and EG = DG      …..(1)
For triangle ECD,
F and H are the mid-point of CD and EC.
Therefore HF || DE and 
HF = `[1]/[2]` DE                                          ....(2)

From (1) and (2) we get,
HF = EG and HF || EG
Similarly, we can show that EH = GF and EH || GF
Therefore GEHF is a parallelogram.