Question

prove that the diagonal of parallelogram bisect each other......​

Answer 1

$\huge\sf \orange{hello}..$

Let consider a parallelogram ABCD in which AB||CD and AD||BC.

In ∆AOB and ∆COD , we have

∠DCO=∠OAB (ALTERNATE ANGLE)

∠CDO= ∠OBA. (ALTERNATE ANGLE)

AB=CD. (OPPOSITE SIDES OF ||gram)

therefore , ∆ AOB ≅ ∆COD. (ASA congruency)

hence , AO=OC and BO= OD. (C.P.C.T)

Answer 2

Given :-

A parallelogram ABCD such that its diagonals AC and BD intersect at O.

To Prove :-

OA = OC and OB = OD

Proof :-

Since ABCD is a parallelogram. Therefore,

AB || DC and AD || BC

AD || BC and transversal DB intersects them at D and B respectively.

∴ ∠CBD = ∠ADB [∴ Alternate interior angles are equal ]

$\implies$ ∠CBO =∠ADO ....1)

Again, AD || BC and AC intersects them at A and C respectively.

∴ ∠ACB = ∠DAC [∴ Alternate interior angles are equal ]

$\implies$ ∠ACO = ∠DAO ....2)

In ∆s AOD and BOC,

∠CBO =∠ADO [from (1)]

AD = BC [ opposite sides of ||gm are equal ]

and,

∠ACO = ∠DAO [from (2)]

∆AOD ≅ ∆BOC, [By ASA]

$\implies$ OA = OC and OB = OD [∴ by Corresponding part of congruent triangle]

Hence, OA = OC and OB = OD

Thanks....