prove that diagonal of a rectangle or parallelogram divides it into two congruent triangles
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Answer:
In ΔBAC and ΔDCA
1 = 2
3 = 4 [alternate pair]
AC = AC (common)
Therefore, diagonal of a parallelogram divides it into two congruent triangles.
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Step-by-step explanation:
Given :
A Parallelogram ABCD and AC is its Diagonal
Prove :
∆ ABC = ∆ CDA
Proof :
In ∆ ABC and ∆ CDA
< DAC = < BCA ( alternate Interior Angle )
So AD || BC
AC = AC ( Common side )
< BAC = < DAC ( alternate Interior Angle )
So AD || BC
< BAC = < DAC ( alternate Interior Angle )
So AB || DC
Therefore By ASA Congruence axiom
So ∆ ABC Congruent ∆ CDA
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