Parallelograms Theorem 5


Ask a Question, Get an Answer!
Hundreds of tutors are online and ready to help you right now!

Statement

The diagonals of a rhombus are perpendicular to each other.

Given:

ABCD is a rhombus. Diagonal AC and BD intersect at O.

To prove:

AC and BD bisect each other at right angles.

Proof:

A rhombus is a parallelogram such that

AB=DC=AD=BC ---(i)

Also the diagonals of a parallelogram bisect each other.

Hence BO=DO and AO=OC ---(ii)

Now compare triangles AOB and AOD,

AB=AD (from (i) above)

BO=DO (from (ii) above)

AO=AO (common side)

(SSS congruency condition)

(corresponding parts of corresponding parts)

BD is a straight line segment.

i.e., the diagonals bisect at right angles.

Hence the theorem is proved.



Ask a Question? Get an Answer!

connect to a tutor


Related Searches

parallelogram proof bisecting line

;,  

theorem5

,  

converse of theorem 5

,  

rhombus

...more