The figure below shows angle cad and angle bisector ae. line segments ed and ec drawn to form aed and aec. points c and d are equidistant from point a. complete the following reason for statement 5 to prove that aec ≅ aed.
The figure below shows angle cad and angle bisector ae. line segments ed and ec drawn to form aed and aec. points c and d are equidistant from point a. complete the following reason for statement 5 to prove that aec ≅ aed.
Here is a picture of the completed table for this answer. Hope this helps.
[tex]The figure below shows angle cad and angle bisector ae. line segments ed and ec drawn to form aed an[/tex]
Me: 2. AD ≅ AC; is equidistant from point AMe: 3. < CAE ≅ < DAE; AE is the bisectorMe: 4. AE ≅ AE; Reflexive Property
NOTE: "<" means angle, and "^" means triangle
Since ray AE is the angle bisector of < CAD, by definition the angle is "bisected into two equal halves of the original angle."
Therefore < CAE ≅ < DAE by the angle bisection law. So now the triangles formed by the bisector, ^CAE and ^DAE, have the two angles just mentioned as the same; also, they share AE as a side.
Since they share the same side, then those sides are ≅ for both ^CAE and ^DAE.
Next, it stated that "Points C and D are equidistant from point A."
Equidistant is a fancy way of saying they are the same length, that side AC ≅ side AD.
NOW WE HAVE WHAT WE NEED!!
By the Side-Angle-Side theorem of triangles, called SAS for short, two triangles are congruent (≅) if two sides of each ^ are ≅ to each other, as well as the < of each that lies in between those two sides of each ^
Therefore, by SAS proving ^CAE ≅ ^DAE, congruent triangles by rule have the other side AND the other 2 angles congruent in each.
So that proves the statement (5) that <AEC ≅ <AED.
**sorry so long but proof need to be accurate step-by-step. 🙂