# The figure below shows angle cad and angle bisector ae. line segments ed and ec drawn to form aed and

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.

## This Post Has 3 Comments

1. whocaresfasdlaf9341 says:

Here is a picture of the completed table for this answer.  Hope this helps.

$The figure below shows angle cad and angle bisector ae. line segments ed and ec drawn to form aed an$

2. toni240 says:

Me: 2. AD ≅ AC; is equidistant from point AMe: 3. < CAE ≅ < DAE; AE is the bisectorMe: 4. AE ≅ AE; Reflexive Property

3. zafarm2oxgpmx says:

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. 🙂