Notifications
Clear all
Quadrilaterals
1
Posts
2
Users
0
Likes
217
Views
0
13/07/2021 11:33 am
Topic starter
Show that the diagonals of a square are equal and bisect each other at right angles.
1 Answer
0
13/07/2021 11:34 am
Let ABCD be a square and its diagonals AC and BD intersect each other at O.
To show that,
AC = BD
AO = OC
and ∠AOB = 90°
Proof:
In ΔABC and ΔBAD,
AB = BA (Common)
∠ABC = ∠BAD = 90°
BC = AD (Given)
ΔABC ≅ ΔBAD [SAS congruency]
AC = BD [CPCT]
diagonals are equal.
In ΔAOB and ΔCOD,
∠BAO = ∠DCO (Alternate interior angles)
∠AOB = ∠COD (Vertically opposite)
AB = CD (Given)
ΔAOB ≅ ΔCOD [AAS congruency]
Thus,
AO = CO [CPCT].
Diagonal bisect each other.
In ΔAOB and ΔCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
ΔAOB ≅ ΔCOB [SSS congruency]
∠AOB = ∠COB
∠AOB + ∠COB = 180° (Linear pair)
Thus, ∠AOB = ∠COB = 90°
Diagonals bisect each other at right angles