Question

Prove that the diagonal of a parallelogram bisects each other .

Given A//gm ABCD is a quadrilateral where AB//CD and BC//AD and AC and BD intersect at O point .

Prove OA = OC and OB = OD

In ∆AOB congruent ∆COD we have

AB = CD (opposite side)
<OAB = <OCD (alternate interior)
<OBA = <ODC (alternate interior)

therefore ∆AOB congruent ∆COD [by ASA congruence]
Hence OA = OC and OB = OD [CPCT]
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Answer 1

Answer:

If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Answer:

Class_9_Maths_Quadrilaterals_Rectangle1

Let ABCD be a parallelogram. To show that ABCD is a rectangle, we have to prove that one of

its interior angles is 900.

In ΔABC and ΔDCB,

AB = DC (Opposite sides of a parallelogram are equal)

BC = BC (Common)

AC = DB (Given)

By SSS congruence rule,

ΔABC ≅ ΔDCB

So, ∠ABC = ∠DCB

It is known that the sum of measures of angles on the same side of traversal is 1800

∠ABC + ∠DCB = 1800 [AB || CD]

=> ∠ABC + ∠ABC = 1800

=> 2∠ABC = 1800

=> ∠ABC = 900

Since ABCD is a parallelogram and one of its interior angles is 900, ABCD is a rectangle.