Question

Prove by vector method that a quadrilateral is a square if and only if its diagonals are congruent and bisect each other at right angles.

Answer 1

Hint: We show two triangles having diagonals as their respective sides congruent using SAS congruence rule. Use the triangle law of vector addition and write diagonals in the form of sides of squares. We show diagonals bisect at right angles by multiplying the diagonal vectors to get the product as zero.
* SAS congruence rule is Side angle side congruence rule where the adjacent sides and the angle between the adjacent sides of one triangle is equal to the adjacent sides and the angle between the adjacent sides of the other triangle.
* Two vectors are said to be perpendicular to each other if their product is zero.
* If in a\[\vartriangle ABC\], sides are represented as vectors\[\overrightarrow {AB} ,\overrightarrow {BC} ,\overrightarrow {AC} \], then sum of vectors of two sides in same direction of triangle is equal to vector of third side of triangle i.e. \[\overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC} \]

Complete step-by-step answer:
Given: A square ABCD
To prove: Diagonals are congruent and bisect at right angles.
Proof:
We draw a square ABCD having sides \[\overrightarrow {AB} ,\overrightarrow {BC} ,\overrightarrow {CD} ,\overrightarrow {DA} \] and diagonals \[\overrightarrow {AC} ,\overrightarrow {DB} \] intersecting at point O.

We have ABCD as a square, which means all angles are right angles and all sides are of equal magnitude.
\[\left| {\overrightarrow {AB} } \right| = \left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {CD} } \right| = \left| {\overrightarrow {DA} } \right|\] and \[\angle A = \angle B = \angle C = \angle D = {90^ \circ }\]
We know that magnitudes of the same vector in both directions are equal
\[\left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {CB} } \right|\]
\[\left| {\overrightarrow {AD} } \right| = \left| {\overrightarrow {DA} } \right|\]
Now we will consider two triangles, \[\vartriangle ABC,\vartriangle BAD\]
Using triangle law of vectors in \[\vartriangle ABC\]
\[\overrightarrow {AB} + \overrightarrow {BC} = \overrightarrow {AC} \] {as vectors AB and BC are in same direction and AC is in different direction}
Since vectors BC and AD are in same direction, we substitute\[\overrightarrow {BC} = \overrightarrow {AD} \]
\[\overrightarrow {AB} + \overrightarrow {AD} = \overrightarrow {AC} \] … (1)
Using triangle law of vectors in \[\vartriangle BAD\]
\[\overrightarrow {DA} + \overrightarrow {AB} = \overrightarrow {DB} \] {as vectors DA and AB are in same direction and BD is in different direction}
 Since vectors DA and AD are in opposite direction, we substitute\[\overrightarrow {DA} = - \overrightarrow {AD} \]
\[\overrightarrow {AB} - \overrightarrow {AD} = \overrightarrow {DB} \] … (2)
We prove congruence of two triangles
\[\left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {DA} } \right|\] (Sides of a square)
\[\angle A = \angle B\]
\[\left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {BC} } \right|\] (Common base in both triangles)
So, by SAS (side angle side) convergence rule which states that two triangles are congruent to each other if two sides and the included angle of one triangle are equal to the corresponding two sides and the included angle of another triangle.
\[ \Rightarrow \vartriangle ABC \cong \vartriangle BAD\]
When two triangles are congruent, we can write their corresponding sides and angles congruent.
Therefore,\[\overrightarrow {AC} \cong \overrightarrow {DB} \]
So the diagonals are congruent.
Now we have to show that diagonals bisect each other at right angles.
From equations (1) and (2) we have
\[\overrightarrow {AB} + \overrightarrow {AD} = \overrightarrow {AC} \]
\[\overrightarrow {AB} - \overrightarrow {AD} = \overrightarrow {DB} \]
We have to show the vector product of the two vectors is zero.
\[ \Rightarrow \overrightarrow {AC} \overrightarrow {DB} \times = (\overrightarrow {AB} + \overrightarrow {AD} ) \times (\overrightarrow {AB} - \overrightarrow {AD} )\]
\[ \Rightarrow \overrightarrow {AC} \times \overrightarrow {DB} = \overrightarrow {AB} \times \overrightarrow {AB} - \overrightarrow {AB} \overrightarrow {AD} + \overrightarrow {AD} \overrightarrow {AB} - \overrightarrow {AD} \overrightarrow {AD} \]
Cancel terms with opposite signs
\[ \Rightarrow \overrightarrow {AC} \times \overrightarrow {DB} = \overrightarrow {AB} \times \overrightarrow {AB} - \overrightarrow {AD} \overrightarrow {AD} \]
Use the property of vectors \[\overrightarrow a \overrightarrow a = {\left| {\overrightarrow a } \right|^2}\]
\[ \Rightarrow \overrightarrow {AC} \overrightarrow {DB} = {\left| {\overrightarrow {AB} } \right|^2} - {\left| {\overrightarrow {AD} } \right|^2}\]
We know \[\left| {\overrightarrow {AB} } \right| = \left| {\overrightarrow {AD} } \right|\] because they both are sides of square having the same magnitude.
\[ \Rightarrow \overrightarrow {AC} \overrightarrow {DB} = {\left| {\overrightarrow {AB} } \right|^2} - {\left| {\overrightarrow {AB} } \right|^2}\]
\[ \Rightarrow \overrightarrow {AC} \overrightarrow {DB} = 0\]
So, \[\overrightarrow {AC} \bot \overrightarrow {DB} \]
So, the diagonals bisect each other at right angles.
Hence Proved

Note: Students might make mistakes when writing the equation for diagonals. Keep in mind that the vectors having the same magnitude but different directions are represented by opposite signs (plus and minus) before them.