Lesson Proof: The diagonals of parallelogram bisect each other

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In this lesson we will prove the basic property of parallelogram in which diagonals bisect each other.

Theorem If ABCD is a parallelogram, then prove that the diagonals of ABCD bisect each other.


drawing( 160, 160, -10, 10, -10, 10, line( -6, -6,4,-6) , line( -6, -6,-2,2), line( -2, 2,8,2) , line( 8, 2,4,-6),line( -6, -6,8,2) ,line( -2, 2,4,-6) ,locate( -6.5,-6.5,A),locate( 4.5,-6.5,B),locate( 8.5,3.5,C),locate(-2.5,4,D),locate(-0,-2.4,O))


Proof

Let the two diagonals be AC and BD and O be the intersection point.

We have to prove that O is the midpoint of AC and also the midpoint of BD.

Hence, AO=OC and BO=OD

We will prove using congruent triangles concept.

Consider two Triangles ABO and COD.

1.  angle OAB = angle CAB = angle ACD = angle OCD ....( Line AC is a transversal of the parallel lines AB and CD, hence alternate angles).

2. angle ODC = angle BDC = angle DBA = angle OBA ....(Line BD is a transversal of the parallel lines AB and CD, hence alternate angles).

3. angle DOC = angle AOB ....(Opposite angles when two lines intersect each other area equal)

From conditions 1,2 and 3

Triangle ABO is similar to triangle CDO (By Angle -Angle similar property)

Since Triangles are similar, Hence ratio of sides are equal from similar triangles property.

(DC/AB)=(DO/OB)=(CO/OA) .........(4)

From theorem that Opposite sides of a parallelogram are equal,

DC=AB ..........(5)

From equation (4) and (5)

(DC/AB)=(DO/OB)=(CO/OA)=1

DO/OB = 1

DO = OB

Similarly, CO=OA

Hence, We conclude that AO = CO and BO = DO.

QED

To learn more about Similar and congruent triangles you can refer to wikipedia article.

To learn more about parallelogram you can refer to wikipedia

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