show that the digonals of square are equal and bisect each other at right angle
Answers
ANSWER:-
Given:
Show that the diagonals of square are equal and bisect at right angle
To prove:
⏺️AC = BD
⏺️AC & BD bisect each other at right angle.
Proof:
In ∆ABC & ∆BAD
AB= BA [common]
BC= AD [opposite sides of sq. ABCD]
∠ABC = ∠BAD [each 90°]
Therefore,
ABCD is a square.
∆ABC ≅∆BAD [congruence rule SAS]
So,
AC = BD [c.p.c.t]
In ∆OAD & ∆OCB,
AD = CB [opposite sides of sq.ABCD]
∠OAD = ∠OCB
AD||BC & transversal AC intersect them
∠ODA = ∠OBC
Therefore,
AD||BC & transversal BD interacts them
∆OAD≅∆OCB [ASA congruence rule]
So,
OA = OC.........(1)
similarly, we can prove that,
OB= OD..........(2)
In equation (1) & (2), we get;
AC & BD bisect each other.
Again,
In ∆OBA & ∆ODA
OB= OD [from (2)]
BA = DA [opposite sides of sq. ABCD]
OA = OA [Common]
Therefore,
∆OBA ≅∆ODA [SSS congruence rule]
∠AOB = ∠AOD [c.p.c.t]
So,
∠AOB +∠AOD = 180° [Linear pair]
∠AOB= ∠AOD = 90°
Therefore,
AC & BC bisect each other at right angles.
Hope it helps ☺️
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.
