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Prove that a diagonal of a parallelogram...

Prove that a diagonal of a parallelogram divides it into two congruent triangles.

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opposites sides of parallelogram are parallel. Thus
So, `AB||DC` and `AD||BC`.
In `/_\ABC` and `/_\ADC`,
`/_BAC=/_DCA`
`AC=AC` (common)
`/_DAC=/_BCA`
Thus, with ASA congruency, `/_\ABC` and `/_\ADC` are congruent.
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Prove that each of the following diagonals of a parallelogram divides it into two congurent triangles. The following lowing are the steps involved in proving the above results. Arrange them in sequential order. ltimg src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/PS_MATH_VIII_C16_E04_031_Q01.png" width="80%"gt (A) By SSS conguruence property , Delta DAB ~= Delta BCD . (B) Let ABCD be a parallelogram and join BD. (C ) AB=CD,AD=BC (opposite sides of parallelogram) and BD =BD (common side). (D ) Similarly , AC divides the parallelogram into two congruent triangles.

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  • Prove that each of the diagonals of a parallelogram divides it into two congruent triangles .The following steps are involved in proving the above result .Arrange them in sequential order . (A) By SSS congruence property , DeltaDAB~=DeltaBCD . (B) Let ABCD be a parallelogram and join BD . ( C) AB=CD,AD =BC (opposite sides of the parallelogram ), and BD =BD(common side) (D) Similarly ,AC divides the parallelogram into two congruent triangles .

    A
    ABCD
    B
    BCAD
    C
    BACD
    D
    CBAD

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