Question

There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is congruent to side CD. If ∠BAC ≅ ∠DCA, the Choose... theorem can be used to show that △ABE ≅ △CDE.

Answer 1

Answer:

SAS theorem

Step-by-step explanation:

Given

$\square ABCD$

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$

$\angle BAC = \angle DCA$

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

$\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$

and

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$

So, the congruent sides and angles of △ABE and △CDE are:

$\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD$ ---- S

$\angle BAC = \angle DCA$ ---- A

$\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED$ or $\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC$  --- S

Hence, the theorem that compares both triangles is the SAS theorem