ABCD is a parallelogram. E and F are tha mid pts of the sides AB and CD respectively. The straight lines AF and BF meet the st. lines ED and EC in poins G and H respectively. Prove that (i) triangle HEB is congruent to triangle FHC
MID PTS THEOREM
in the triangles FHC and HEB
EB=FC....since they are half parts of equal sides AB ,CD.
FC II EB
angle FCH= angle HEB
angle CFH= angle HBE...(alt interior angles)
so triangles are cong by ASA rule.
consider the quadrilaterlal AECF,
AE=CF....half parts of equal sides AB and CD
AE II CF...since AB , CD are parallel.
so AECF is a parallelogram.
GF II EH
GE II HF
so GEHF is also a parallelogram... both pairs of opp sides are parallel.