Question

13.Show that if the diagonals of a Quadrilateral are equal and bisect each other at right angles,then it is a Square.​

Answer 1

$\displaystyle\large\underline{\sf\red{Given}}$

✭ Diagonals of a quadrilateral are equal

✭ They bisect each other at 90°

$\displaystyle\large\underline{\sf\blue{To \ Prove}}$

◈ It is a square?

$\displaystyle\large\underline{\sf\gray{Solution}}$

So here we may show that two triangles are congruent and then show that the sides and the angles are equal!!

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$\underline{\bigstar\:\textsf{According to the given Question :}}$

In ∆AOB & ∆COD

➝ $\displaystyle\sf AO = CO \:\{Diagonals \ Bisect \ each \ other\}$

➝ $\displaystyle\sf \angle AOB = \angle COD \:\{Vertically \ Opposite\}$

➝ $\displaystyle\sf OB = OD \:\{Diagonals \ Bisect \ each \ other\}$

$\displaystyle\sf \therefore \purple{\triangle AOB \cong \triangle COD \: \{SAS\}}$

So from that we may see that AB = CD [CPCT]

In ∆AOD & ∆COD

»» $\displaystyle\sf AO = CO \:\{Diagonals \ bisect \ each \ other\}$

»» $\displaystyle\sf \angle AOD = \angle COD \:\{Vertically \ Opposite\}$

»» $\displaystyle\sf OD = OD \:\{Common\}$

$\displaystyle\sf \therefore \green{\triangle AOD \cong \triangle COD\:\{SAS\}}$

So from this we may see that AD = DC [CPCT]

In ∆AOD & ∆BOC

➢ $\displaystyle\sf DO = BO \: \{Diagonals \ bisect \ each \ other\}$

➢ $\displaystyle\sf \angle AOD = \angle BOC \:\{Vertically \ Opposite\}$

➢ $\displaystyle\sf AO = OC \: \{Diagonals \ bisect \ each \ other\}$

$\displaystyle\sf \therefore{\orange{\triangle AOD \cong \triangle BOC\:\{SAS\}}}$

So from this we may see that AD = BC

From these observations we may infer that,

⪼ $\displaystyle\sf AB = BC = CD = DA$

Now we shall prove that angle is equal to 90°

So,

✪ $\displaystyle\sf \angle ADC = \angle BCD \:\{CPCT\}$

›› $\displaystyle\sf \angle ADC +\angle BCD = 180^{\circ} \:\{Co-Interior \ Angle\}$

›› $\displaystyle\sf 2\angle ADC = 180^{\circ}$

›› $\displaystyle\sf \angle ADC = \dfrac{180}{2}$

›› $\displaystyle\sf \pink{\angle ADC = 90^{\circ}}$

So we know that any figure who's sides are equal and Angles are equal to 90° is a Square

$\displaystyle\sf \sf\star \: Diagram \:\star$

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(4,1)\qbezier(1,1)(1,1)(1,4)\qbezier(1,4)(1,4)(4,4)\qbezier(4,4)(4,4)(4,1)\qbezier(1,1)(1,1)(4,4)\qbezier(4,1)(4,1)(1,4)\put(0.8,0.5){\sf D}\put(0.8,4.3){\sf A}\put(4,4.3){\sf B}\put(4,0.5){\sf C}\put(2.4,2){\sf O}\end{picture}$

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Answer 2

Step-by-step explanation:

Explanation:

______________________________

Given that,

Let ABCD be a quadrilateral

It's iagonals AC and BD bisect each other at right angle at O.

To prove that

The Quadrilateral ABCD is a square.

Proof,

In ΔAOB and ΔCOD,

⇝ AO = CO (Diagonals bisect each other)

⇝ ∠AOB = ∠COD (Vertically opposite)

⇝ OB = OD (Diagonals bisect each other)

⇝ ΔAOB ≅ ΔCOD [SAS congruency]

Thus,

⇝ AB = CD [CPCT] — (i)

also,

∠OAB = ∠OCD (Alternate interior angles)

⇒ AB || CD

Now,

⇝ In ΔAOD and ΔCOD,

⇝ AO = CO (Diagonals bisect each other)

⇝ ∠AOD = ∠COD (Vertically opposite)

⇝ OD = OD (Common)

⇝ ΔAOD ≅ ΔCOD [SAS congruency]

Thus,

AD = CD [CPCT] ____ (ii)

also,

AD = BC and AD = CD

⇒ AD = BC = CD = AB ____ (ii)

also, ∠ADC = ∠BCD [CPCT]

and ∠ADC + ∠BCD = 180° (co-interior angles)

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90° ____ (iii)

One of the interior angles is right angle.

Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

HenceProved!