It is demonstrated that a parallelogram can be divided into two congruent triangles by a diagonal.
Given:
Parallelogram = BKJI
The line segments BI and KJ are parallel to each other and BJ is the transversal or the diagonal of a parallelogram.
Diagonal = BJ
We have to prove that the Diagonal of a parallelogram divides it into two congruent triangles.
The parallelogram is divided into two parts which are triangles BJK and BJI.
In ΔBJK and ΔBJI
BI || KJ and BJ is a transversal
Thus,
∠ BJK = ∠ JBI (vertically opposite angles are always equal)
BJ = BJ (common side)
So, similarly ∠JBK = ∠ BJI (vertically opposite angles)
Thus, by ASA congruency ΔBJK ≅ ΔBJI
Hence the two triangles are congruent.
To learn more about congruency of triangles visit:
brainly.com/question/20521780
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