To prove that ΔAED ˜ ΔACB by SAS, Jose shows that StartFraction A E Over A C EndFraction = StartFraction A D Over A B EndFraction.

Triangle A E D is shown. Line segment B C is drawn from side A D to A E to form triangle A C B.

Jose also has to state that


please help!!!!!!

Respuesta :

Answer:

A on Edg.

Step-by-step explanation:

Took the Test

bec97

Answer:

<A is congruent to <A

Step-by-step explanation:

<A is in both triangle and it can be equal to itself because of the reflexive property.