show that the diagonals of a square are equal and bisect each other at right angles
Answers
Answer:
HEY MATE HERE IS YOUR ANSWER
Step-by-step explanation:
GIVEN:
ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.
RTP:
the diagonals of a square are equal and bisect each other at right angles,
AC = BD, OA = OC, OB = OD, and ∠AOB = 90º.
PROOF:
- In ΔABC and ΔDCB,
AB = DC (Sides of a square are equal to each other)
∠ABC = ∠DCB (All interior angles are of 90)
BC = CB (Common side)
So, ΔABC ≅ ΔDCB (By SAS congruency)
Hence, AC = DB (By CPCT)
Hence, the diagonals of a square are equal in length.
- In ΔAOB and ΔCOD,
∠AOB = ∠COD (Vertically opposite angles)
∠ABO = ∠CDO (Alternate interior angles)
AB = CD (Sides of a square are always equal)
So, ΔAOB ≅ ΔCOD (By AAS congruence rule)
Hence, AO = CO and OB = OD (By CPCT)
Hence, the diagonals of a square bisect each other.
- In ΔAOB and ΔCOB,
As we had proved that diagonals bisect each other, therefore,
AO = CO
AB = CB (Sides of a square are equal)
BO = BO (Common)
So, ΔAOB ≅ ΔCOB (By SSS congruency)
Hence, ∠AOB = ∠COB (By CPCT)
However,
∠AOB + ∠COB =180°{Linear Pair}
AOB+AOB=180°
*2<AOB=180°
<AOB=180°/2
<AOB=90°
Hence PROVED....
{/* ANGLE /}
Step-by-step explanation:
Given that ABCD is a square.
To prove : AC=BD and AC and BD bisect each other at right angles.
Proof:
(i) In a ΔABC and ΔBAD,
AB=AB ( common line)
BC=AD ( opppsite sides of a square)
∠ABC=∠BAD ( = 90° )
ΔABC≅ΔBAD( By SAS property)
AC=BD ( by CPCT).
(ii) In a ΔOAD and ΔOCB,
AD=CB ( opposite sides of a square)
∠OAD=∠OCB ( transversal AC )
∠ODA=∠OBC ( transversal BD )
ΔOAD≅ΔOCB (ASA property)
OA=OC ---------(i)
Similarly OB=OD ----------(ii)
From (i) and (ii) AC and BD bisect each other.
Now in a ΔOBA and ΔODA,
OB=OD ( from (ii) )
BA=DA
OA=OA ( common line )
ΔAOB=ΔAOD----(iii) ( by CPCT
∠AOB+∠AOD=180° (linear pair)
2∠AOB=180°
∠AOB=∠AOD=90°
∴AC and BD bisect each other at right angles.
