The diagonals of parallelogram ABCD meets at O- Then - COD congruent to

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Standard IX

Mathematics

Theorem 1: Diagonal of a Parallelogram Divides It into Two Congruent Triangles

The diagonals...

Question

The diagonals of parallelogram ABCD meets at O. Then
$Δ C O D$ congruent to

(\ \Delta COB\)

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(\ \Delta AOD\)

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(\ \Delta AOB\)

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Solution

The correct option is C (\ \Delta AOB\)

AB = CD (opposite sides of parallelogram)

$∠ O C D = ∠ O A B$
$∠ O D C = ∠ O B A$ (alternate interior angles)

$T h e r e f o r e$ by ASA postulate $Δ C O D ≅ Δ A O B$

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\frac{Area (∆ AOB)}{Area (∆ COD)}

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meeting

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The diagonals of parallelogram ABCD meets at O. Then ΔCOD congruent to

The correct option is C (\ \Delta AOB\) AB = CD (opposite sides of parallelogram) ∠OCD=∠OAB ∠ODC=∠OBA (alternate interior angles) Therefore by ASA postulate ΔCOD≅ΔAOB

The diagonals of parallelogram ABCD meets at O. Then
$Δ C O D$ congruent to

The correct option is C (\ \Delta AOB\)

AB = CD (opposite sides of parallelogram)

$∠ O C D = ∠ O A B$
$∠ O D C = ∠ O B A$ (alternate interior angles)

$T h e r e f o r e$ by ASA postulate $Δ C O D ≅ Δ A O B$