prove that the diagonals divide the llgm into 2 congruent triangles
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Consider a parallelogram ABCD
Diagonal- BD
In ∆ABD and ∆BCD,
(Alternate angles)
BD = BD (common)
Therefore, ∆ABD =~ ∆BCD by SAS
Diagonal- BD
In ∆ABD and ∆BCD,
(Alternate angles)
BD = BD (common)
Therefore, ∆ABD =~ ∆BCD by SAS
Answered by
2
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