Question

The diagonal of a parallelogram bisect each other prove.​

Answer 1

We have to prove that the diagonals bisect each other in a parallelogram ABCD.

∠ABE≃∠CDE ( by alternate interior angles )

∠DEC≃∠BAE ( alternate interior angles )

∠ABE≃∠CDE by ASA

⇒{AE=CEBE=CE} coordinate sides of ≅ triangle

∠CED≅∠BEA [ vertical angles ]

△AEB≅△DEC [ by SAS≅A

′

s ]

∠CDE≅∠BAE [ coordinate angles of ≅ triangles. ]

∴AB∥CD by alternate interior angles ≅ of parallel lines.

∠AEC≅∠DEB ( vertical angles )

△AEC≅△DEB ( by SAS )

∠CAE≅∠BDE [ coordinate angle ]

∴AC∥BC by alternate interior angles

Theorm :

A quadrilateral is a parallelogram if and only if the diagonals bisect each other.

Hence, the answer proved.

Answer 2

Answer :-

Let the parallelogram be ABCD, whose diagnals are AC and BD, which intersects at O.

Required figure :

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf A}\put(6,0.5){\sf B}\put(1.4,4.3){\sf D}\put(6.6,4.3){\sf C}\put(3.7,2.1){\sf O}\end{picture}$

Theorem : The diagnals of a parallelogram bisect each other.

To Prove : The diagnals of parallelogram bisect each other ( OA = OC and OB = OD )

Proof : In ∆s OAB and OCD, we have,

∠ OAB = ∠ OCD ........ [ Alternate angles ]

∠ OBA = ∠ ODC ....... [ Alternate angles ]

AB = CD ......... [ Opposite sides of the parallelogram ]

Therefore, ∆OAB = ∆OCD ....... [ c.p.c.t. ]

Hence,

OA = OC and OB = OD ...... [ c.p.c.t. ]

Hence, Proved !