prove that in a square diagonals bisect each other
Answers
Step-by-step explanation:
Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O. To prove that the diagonals of a square are equal and bisect each other at right angles, we have to prove AC = BD, OA = OC, OB = OD, and AOB = 90º.
In ABC and DCB,
AB = DC (Sides of a square are equal to each other)
ABC = DCB (All interior angles are of 90)
BC = CB (Common side)
ABC = DCB (By SAS congruency)
AC = DB (By CPCT)
Hence, the diagonals of a square are equal in length.
In AOB and COD,
AOB = COD (Vertically opposite angles)
ABO = CDO (Alternate interior angles)
AB = CD (Sides of a square are always equal)
AOB = COD (By AAS congruence rule)
AO = CO and OB = OD (By CPCT)
Hence, the diagonals of a square bisect each other.
In AOB and COB,
As we had proved that diagonals bisect each other, therefore,
AO = CO
AB = CB (Sides of a square are equal)
BO = BO (Common)
AOB = COB (By SSS congruency)
AOB = COB (By CPCT)
However,AOB + COB = 180 (Linear pair)
2 AOB = 180º
AOB = 90º
Hence, the diagonals of a square bisect each other at right angles.