Answer:
SAS theorem
Step-by-step explanation:
Given
[tex]\square ABCD[/tex]
[tex]\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD[/tex]
[tex]\angle BAC = \angle DCA[/tex]
Required
Which theorem shows △ABE ≅ △CDE.
From the question, we understand that:
AC and BD intersects at E.
This implies that:
[tex]\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC[/tex]
and
[tex]\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED[/tex]
So, the congruent sides and angles of △ABE and △CDE are:
[tex]\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD[/tex] ---- S
[tex]\angle BAC = \angle DCA[/tex] ---- A
[tex]\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED[/tex] or [tex]\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC[/tex] --- S
Hence, the theorem that compares both triangles is the SAS theorem
