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. ∠GAH ≅ ∠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

I am going to label intersection of AG BH as point Z

on proof,

(1) given

(2) BC parallel to AD

(3) alternate interior angles of traverse line through parallel lines

(4) given

(5) HAG=BAG and GBH=ABH angles

(6) equality substitution

(7) BAG and BHA are isosceles triangles

here I start to differ from proof outline in question

(8) BG=BA and BA=AH opposite sides of equal angles in isosceles triangle

now have BG=BA=AH

(9) triangles BGZ and AZH congruent ,,,AAS reason

(10) BZ=ZH and GZ=AZ from congruent triangles

(11) GZH=BZA angles,,,vertical angle reason

(12) construct GH

(13) triangles GZH and AZB congruent,,,SAS reason

(14) GH=AB congruent triangles

(15) AB=BG=AH=GH meaning ABGH is rhombus by definition of rhombus