Question

Prove that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.

Answer 1

Given:- ABCD is a rectangle i.e AB = CD and AD = BC

To Prove:- ABCD is a square only if its diagonal are perpendicular

Proof:- Let A be at the origin

\(\vec a\) and  \(\vec b\) be position vector of B and D respectively

Now,

By parallelogram law of vector addition,

 ⇒ \(\vec{AC}=\vec{AB}+\vec{BC}\)

Since in rectangle opposite sides are equal BC = AD

Negative sign as vector is opposite

Diagonals are perpendicular to each other only 

Hence all sides are equal if diagonals are perpendicular to each other

ABCD is a square

Hence proved

Answer 2

Given: ABCD is a square. AC and BD are diagonals intersecting at O.

To prove: AC=BD and AC⟂BD Proof: AB=AD (sides of a square are equal) AB||DC (opposite sides of a square are parallel)

ABCD is parallelogram with consecutive sides equal. ABCD is a rhombus. (by definition) Since the diagonals of a rhombus are perpendicular to each other, AC⟂BD ABCD is a parallelogram. ABCD is a rectangle with a pair of its consecutive sides equal. Since the diagonals of a rectangle are equal, AC=BD. So, Diagonal AC=Diagonal BD and AC ⟂BD

 Hence the theorem is proved.