Prove that the diagonal of a parallelogram divides it into two congruent triangles
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here is your answer dear..
Step-by-step explanation:
consider Δ ABC and Δ ACD
Since the line segments AB+CD are parallel
to each other and AC is a transversal
∠ ACB = ∠ CAD.
AC = AC (common side)
∠CAB = ∠ ACD.
Thus, by ASA criteria
ΔABC ≅ ΔACD
The corresponding part of the congruent
triangle are congruent
AB = CD + AD = BC
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