if diagonals of a quadrilateral are equal and bisect each other at right angle then it is square
Answers
TO PROVE - AC = BD,
OA=OC, OB =OD
∠AOB =90°
PROOF -
1) IN ΔABC AND ΔDCB
AB=DC ( EQUAL SIDES OF SQUARE)
∠ABC=∠DCB (90° EACH)
BC = CB(COMMON SIDE)
∴ΔABC ≡ ΔDCB (SAS RULE)
AC=DB ( CPCT)
2) IN ΔAOB AND ΔCOD
∠AOB=∠COD (VERTICALLY OPPOSITE ANGLES )
∠ABO=∠CDO (ALTERNATE INTERIOR ANGLES )
AB = CD ( EQUAL SIDES OF SQUARE )
ΔAOB ≡ ΔCOD ( AAS RULE )
AO=CO AND OB = OD ( CPCT)
3) IN ΔAOB AND ΔCOB
AO = CO(PROVED ABOVE)
AB=CB ( EQUAL SIDES OF SQUARE)
BO=BO (COMMON)
ΔAOB≡ΔCOB ( SSS RULE)
∴∠AOB=∠COB ( CPCT )
∠AOB+∠COB=180° ( LINEAR PAIR )
2∠AOB=180° (SINCE,∠AOB=∠COB)
∠AOB=180°/2
∠AOB=90°
THEREFORE, PROVED THAT IN A QUAD. IF DIAGONALS ARE EQUAL AND THEY PERPENDICULARLY BISECT EACH OTHER THEN THEY THE QUAD. IS A SQUARE.
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!