Question

$\color{purple}{Question}$

Show that if the diagonal of a quadrilateral are equal and bisect each other at right angles, then it is a square

$\color{purple} {Be\:Brainly}$
​

Answer 1

Step-by-step explanation:

$\sf \large \pink{\underline{ {Given} }}$

Let ABCD be the quadrilateral.

Diagonals are equal, i.e., AC = BD ...(1)

& they bisect each other, i.e.

OA=OC & OB=OD ...(2)

At right angles,i.e.,

∠AOB= ∠BOC= ∠COD= ∠ AOD=90° ...(3)

$\sf \large \red{ \underline{To \: prove}}$

ABCD is a square

$\sf \large \purple{ \underline{Solution}}$

・Square is a parallelogram with all sides equal and one angle 90°

・First we will prove ABCD is a parallelogram

・After that we'll prove all sides equal, and one angle equal to 90°

$\sf \: In \triangle AOB \: and \: \triangle \: COB, \\ \sf \: OA = OC \: (from \: ... 2 \:) \\ \sf \angle AOB= \angle COB \: (from \: ...3 \: both \: 90 \degree)\\ \sf \: OB =OB \: (common) \\ \sf \therefore \triangle AOB \cong \triangle COB \\ \therefore \sf AB = CB$

Similarly we can prove

$\sf\triangle AOB \cong \triangle DOA , so \: AB = AD \\ \sf \& \triangle BOC \cong \triangle COD,so \: OCB=DC$$\sf \: So,AB=AD=CB=DC \\ \sf \: Now \: we \: can \: say \: that \\ AB=CD \: \& \: AD=BC$

In ABCD, both pairs of opposite sides are equal,

Hence, ABCD is a parallelogram

Square is a parallelogram with all sides equal and one angle 90° So, we prove one angle 90°

$\sf \: In \: \triangle ABC \: and \triangle \: DCB, \\ \sf AC = BD \: (from..1)\\\sf AB = DC \\ \sf \: BC = CB \: (common) \\ \sf \therefore \triangle ABC \cong \triangle DCB \: \\ \sf \Rightarrow \angle ABC= \angle DCB \: (CPCT) ....(4)$

Now,

$\sf \: AB||C \\ \sf \: \& \: BC \: is \: transversal \: \\ \sf \angle B+ \angle C=180 \degree \\ \sf \: \angle B+ \angle B=180 \degree \: (from \: ...4) \\ \sf \: 2 \angle B=180 \degree \\ \sf \: \angle B= \frac{180}{ 2} =90 \degree$

$\blue{ \sf \: Thus, \: ABCD \: is \: a \: parallelogram \: with} \\ \sf \blue{all \: sides \: equal \: and \: one \: angle \: 90 \degree} \\ \sf \blue{So, \: ABCD \: is \: a \: square}$

Answer 2

Let ABCD be the quadrilateral.

Diagonals are equal, i.e., AC = BD ...(1)

& they bisect each other, i.e.

OA=OC & OB=OD ...(2)

At right angles,i.e.,

∠AOB= ∠BOC= ∠COD= ∠ AOD=90° ...(3)

ABCD is a square

・Square is a parallelogram with all sides equal and one angle 90°

・First we will prove ABCD is a parallelogram

・After that we'll prove all sides equal, and one angle equal to 90°

Similarly we can prove

In ABCD, both pairs of opposite sides are equal,

Hence, ABCD is a parallelogram

Square is a parallelogram with all sides equal and one angle 90° So, we prove one angle 90°

Now,