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The SAS congruence postulate, â–³AEB ≅ â–³DCB.
The diagram shows two triangles, â–³AEB and â–³DCB, where:
∠A≅∠D
AE ≅ DC
EB ≅ CB
BA ≅ BD
We need to prove that â–³AEB ≅ â–³DCB.
There are multiple ways to approach this problem, but one possible solution is to use the Side-Angle-Side (SAS) congruence postulate.
This postulate states that if two triangles have two sides and the included angle between them congruent, then the triangles themselves are congruent.
In this case, we have:
AE ≅ DC (given)
∠A≅∠D (given)
BA ≅ BD (given)
Therefore, by the SAS congruence postulate, â–³AEB ≅ â–³DCB.
Here's another way to think about it:
Imagine cutting triangle AEB along AE and rotating it until AE coincides with DC .
Since AE ≅ DC and ∠A≅∠D, the two triangles will perfectly overlap when you do this.
This means that the corresponding sides and angles of the two triangles must be congruent.
Therefore, â–³AEB ≅ â–³DCB.
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Rewritten by : Jeany
