Question

Prove that diagonals of a rhombus bisect each other at right angles vectors

Answer 1

Let ABCD be a rhombus.

⇒|AB−→−|=|BC−→−|=|DC−→−|=|AD−→−|

The diagonals are AC−→−andDB−→−

Since Rhombus is a parallelogram we know that in a parallelogram

the diagonals bisect each other.

We have to prove that the diagonals bisect perpendicularly.

i.e., we have to prove that AC−→−is⊥toDB−→−

From parallelogram law of addition we know that

AC−→−=AB−→−+AD−→−andDB−→−=AB−→−−AD−→−

To prove that AC−→−is⊥toDB−→−

⇒proveAC−→−.DB−→−=0

AC−→−.DB−→−=(AB−→−+AD−→−).(AB−→−−AD−→−)

We know that (a→+b→).(a→−b→)=|a→|2−|b→|2

⇒AC−→−.DB−→−=|AB−→−|2−|AD−→−|2

Since ABCD is a Rhombus |AB−→−|=|AD−→−|

⇒AC−→−.DB−→−=0

⇒AC−→−is⊥toDB−→−

⇒ the diagonals of a Rhombus are perpendicular bisectors.

Hence proved.