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.

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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. 🙂