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

SAS theorem

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