Given a parallelogram □ABCD, with AD > AB. The bisector of ∠A intersects line BC at G, and the bisector of ∠B intersects line AD at H. Prove that □ABGH is a rhombus.

Question

Fill in the blanks (____) to complete the proof given in the comments

STATEMENT (REASONS)

1. □ABCD is a parallelogram (____)
2. (____) (Definition of parallelogram)
3. ∠GHA ≅ ∠BGA, ∠GBH ≅ ∠BHA (_____)
4. line BC bisector of ∠A, line AD bisector of ∠B (____)
5. (_____) (Definition of Angle bisector)
6. ∠BGA ≅ ∠BAG, ∠BHA ≅ ∠ABH (____)
7. (____) (CITT)
8. BG=AH (____)
9. AG=AG (____)
10. (____) (SAS POSTULATE)
11. (_____) (_____)

Answer 1

An important error is in line 3...GHA ≅ ∠BGA, ∠GBH ≅ ∠BHA. This is not accurate. GHA is obtuse, BGA is acute