Ex 8.1, 5
Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.
Given:
Let ABCD be the quadrilateral.
Diagonals are equal, i.e., AC = BD
& bisect each other, i.e. OA = OC & OB = OD,
at right angles ,i.e., ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°
To prove: ABCD is a square
Proof: Square is a parallelogram with all sides equal and one angle 90°
First we will prove ABCD is a parallelogram
and then prove all sides equal , and one angle equal to 90°
In ΔAOB and ΔCOB,
OA = OC
∠AOB = ∠COB
OB = OB
∴ ΔAOB ≅ ΔCOB
∴ AB = CB
Similarly we can prove
ΔAOB ≅ ΔDOA , so AB = AD
& ΔBOC ≅ ΔCOD , so CB = DC
So, AB = AD = CB = DC
Now we can say that
AB = CD & AD = BC
In ABCD, both pairs of opposite sides are equal,
Hence, ABCD is a parallelogram
Square is a parallelogram with all sides equal and one angle 90°
So, we prove one angle 90°
In ΔABC and ΔDCB,
AC = BD
AB = DC
BC = CB
∴ ΔABC ≅ ΔDCB
⇒∠ ABC = ∠ DCB
Now,
AB ∥ CD
& BC is transversal
∠ B + ∠ C = 180°
∠ B + ∠ B = 180°
2∠ B = 180°
∠ B = (180°)/2 = 90°
Thus, ABCD is a parallelogram with all sides equal and one angle 90°
So, ABCD is a square

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.