Respuesta :

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

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