4. Show that the diagonals of a square are equal and bisect each other at right angles.
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] Thus, AC = BD [CPCT] diagonals are equal. Now, 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. Now, In ΔAOB and ΔCOB, OB = OB (Given) AO = CO (diagonals are bisected) AB = CB (Sides of the square) , ΔAOB ≅ ΔCOB [SSS congruency] also, ∠AOB = ∠COB ∠AOB+∠COB = 180° (Linear pair) Thus, ∠AOB = ∠COB = 90° , Diagonals bisect each other at right angles
