Question

Prove all the converse of the properties of parallelogram

Answer 1

heyy....

There are five ways in which you can prove that a quadrilateral is a parallelogram. The first four are the converses of parallelogram properties (including the definition of a parallelogram). Make sure you remember the oddball fifth one — which isn’t the converse of a property — because it often comes in handy:

If both pairs of opposite sides of a quadrilateral are parallel, then it’s a parallelogram (reverse of the definition).

If both pairs of opposite sides of a quadrilateral are congruent, then it’s a parallelogram (converse of a property).

Tip: To get a feel for why this proof method works, take two toothpicks and two pens or pencils of the same length and put them all together tip-to-tip; create a closed figure, with the toothpicks opposite each other. The only shape you can make is a parallelogram.

If both pairs of opposite angles of a quadrilateral are congruent, then it’s a parallelogram (converse of a property).

If the diagonals of a quadrilateral bisect each other, then it’s a parallelogram (converse of a property).

hope its help...

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Answer 2

$\boxed{{Properties}\\ 1. \: each \:diagonal \: divides \: it \: into \: two \: congruent \: triangles. \\ 2. \: the \: opposite \: sides \: and \: angles \: of \: a \: parallelogram \: are \: equal \: \\ 3. \: the \: diagonals \: of \: parallelogram \: bisect \: each \: other .}$

We can prove the above properties as under :

${\fbox\green{Proof~:}}$

Consider a parallelogram ABCD. Draw its diagonal AC .

Then, in traingle ABC and CDA , We have

∠1 = ∠2

(Alternate angles , AB || DC and AC is the transversal)

∠3 = ∠4

(Alternate angles , AD || BC and AC is the transversal)

AC = AC ( Common )

•°• ∆ABC ≈ ∆CDA

(ASA property of congruence of ∆s. )

⟹ AB = CD and BC = DA

( Corresponding parts of congruent triangles )

Also, ∠B = ∠D

Similarly, by drawing the diagonal BD , we prove that

∆ABD ≈ ∆CDB

⟹ ∠A = ∠C

Hence, properties 1 and 2 are proved .

${\fbox\green{To~proof~property~3~:}}$

Consider parallelogram ABCD and draw its diagonal AC and BD. Let these diagonals intersect each other bat a point O.

Then, in traingle OAB and OCD, we have

AB = CD

(opposite sides of a parallelogram)

∠AOB = ∠COD

(vertically opposite angles)

∠OAB = ∠OCD

(Alternate angles ; AB || DC and transversal AC cuts them. )

•°• ∆OAB ≈ ∆OCD

( AAS property of congruence of triangles. )

⟹ OA = OC and OB = OD

( Corresponding parts of congruent ∆s )

Thus proces the diagonal property of parallelogram , $i.e.$, the diagonal of a parallelogram bisect each other .