Question

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Proove that the diagonals of a parallelogram divide it into four congruent Triangles .​

Answer 1

Answer:

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Proove that the diagonals of a parallelogram divide it into four congruent Triangles .

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We know that diagonals of parallelograms bisect each other. Therefore, O is the mid-point of diagonal AC and BD.

BO is the median in ΔABC. Therefore, BO will divide ΔABC into two triangles of equal areas.

∴ Area (ΔAOB) = Area (ΔBOC) ... (1 )

Also, In ΔBCD, CO is the median. Therefore, median CO will divide ΔBCD into two equal triangles.

Hence, Area (ΔBOC) = Area (ΔCOD) ... (2)

Similarly, Area (ΔCOD) = Area (ΔAOD) ... (3)

From Equations equation (1), (2) and (3) we obtain

rea (ΔAOB) = Area (ΔBOC) = Area (ΔCOD) = Area (ΔAOD)

Therefore, we can say that the diagonals of a parallelogram divide it into four triangles of equal area.

Answer 2

Answer:

If two lines parallel to sides of a parallelogram are constructed concurrent to a diagonal, then the parallelograms formed on opposite sides of that diagonal are equal in area. The diagonals of a parallelogram divide it into four triangles of equal area.

Step-by-step explanation:

according to the angle-side-angle property of triangles, triangles ABC and CDA are congruent. This means that the other two sides of these triangles are equal: AB = CD and BC = AD. Thus, the opposite sides in a parallelogram are equal.