Show the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
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Answer:
Step-by-step explanation
Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O.
It is given that the diagonals of ABCD are equal and bisect each other at right angles.
Therefore, AC = BD, OA = OC, OB = OD, and ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
To prove ABCD is a square, we have to prove that ABCD is a parallelogram in which AB = BC = CD = AD, and one of its interior angles is 90°.
In ΔAOB and ΔCOD,
AO = CO (Diagonals bisect each other)
OB = OD (Diagonals bisect each other)
∠AOB = ∠COD (Vertically opposite angles)
∴ ΔAOB = ΔCOD (SAS congruence rule)
∴ AB = CD (By CPCT) ,,,,,,, (1)
And, ∠OAB = ∠OCD (By CPCT)
However, these are alternate interior angles for line AB and CD and alternate interior angles are equal to each other only when the two lines are parallel.
Thus, AB |/CD ,,,,,,,,,,,,,,,,,(2)
In ΔAOD and ΔCOD,
AO = CO (Diagonals bisect each other)
∠AOD = ∠COD (Each angle is 90°)
OD = OD (Common)
∴ ΔAOD ≅ ΔCOD (SAS congruence rule)
∴ AD = DC (By CPCT) ,,,,,,,,,,,,,,,,,, (3)
However, AD = BC and AB = CD (Opposite sides of parallelogram ABCD are equal)
∴ AB = BC = CD = DA
Therefore, all the sides of quadrilateral ABCD are equal to each other.
In ΔADC and ΔBCD,
AD = BC (Already proved)
AC = BD (Given)
DC = CD (Common)
∴ ΔADC =ΔBCD (SSS Congruence rule)
∴ ∠ADC = ∠BCD (By CPCT)
However, ∠ADC + ∠BCD = 180°(Co-interior angles)
∠ADC + ∠ADC = 180°
2∠ADC = 180°
∴ ∠ADC = 90°
One of the interior angles of quadrilateral ABCD is a right angle.
Thus, we have obtained that ABCD is a parallelogram where AB = BC = CD = AD and one of its interior angles is 90°.
Therefore, ABCD is a square.
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