For the given figure, we will prove the triangles BGH and BDH are congruent.
So, the proof will be as follows:
Statement Reason
1) m∠F = m∠FEG = m∠FGE Given
2) ΔFGE is an equilateral triangle from (1), the definition of an equilateral Δ
3) FG = GE from (2), the definition of equilateral Δ
4) FG = ED Given
5) m∠ GEH = m∠DEH Given
6) EH = EH Reflexive property
7) ΔGEH ≅ Δ DEH SAS postulate
8) GH = DH from (7) CPCTC
9) m∠GHE = m∠DHE from (7) CPCTC
10) m∠GHB = m∠DHB Definition of supplementary angles
11) HB = HB Reflexive property
12) ΔBGH ≅ ΔBDH SAS postualte
13) ΔAGB is an isosceles Δ Given
14)AG = AB Definition of isosceles Δ
15) ΔBCD is an isosceles Δ Given
16) CB = CD Definition of isosceles Δ
17) CB = AG Given
18) m∠CDB = m∠CBD Definition of isosceles Δ
19) m∠ABG = m∠AGB Definition of isosceles Δ
20) m∠CDB = m∠AGB Given
21) m∠ABG = m∠CBD Transitive property
22) Δ AGB ≅ ΔCBD AAS postualte