Answer: d) similar, SAS similarity, ΔDGH
Step-by-step explanation:
DG ≡ GC ⇒ G is the midpoint of DC
DH ≡ HB ⇒ H is the midpoint of DB
Therefore, HB is the midsegment of ΔDCB.
By the Midsegment Theorem, HB || BC
By Corresponding Angles Theorem, ∠DGH ≡ ∠DCB
[tex]\underline{Sides}:\\\\\dfrac{DB}{DC}= \dfrac{7}{14}\rightarrow \dfrac{1}{2}\qquad \dfrac{DH}{DB}= \dfrac{7}{14}\rightarrow \dfrac{1}{2}[/tex]
ΔDCB ≅ ΔDGH by Side-Angle-Side Similarity Theorem
