prove that a diagonal divides a parallelogram into two congruent triangles
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Answered by
1
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
Answered by
1
Step-by-step explanation:
Consider a parallelogram ABCD with diagonal AC.
Now in Triangle ABC & CDA
AB =DC (opposite sides of a parallelogram)
AD=BC (opposite sides of a parallelogram)
AC = AC (common)
Therefore,
by SSS rule,
Triangle ABC congruent to triangle CDA
Hence proved
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