Question

prove that diagonals of square are perpendicular bisector

Answer 1

A square is defined as a rectangle whose adjacent (or all) sides are equal .

Alternately a square is a Rhombus whose adjacent sides are at right angles.
Let O be the intersection of diagonals AC and BD. In triangle ABC, AB=BC. Angle B=90°. So angles BCA and BAC are equal and = 45°.

Similarly in triangle BCD, Angles CBD = angle CDB = 45°. Reason: angle C=90°. BC=CD.

So in triangle BOC we find Angle BOC = 180-45 -45 =90°. So diagonals intersect at right angles.
Also BO = OC as angles OBC = angle OCB. Similarly in triangle AOD we get AO = OD. In triangle AOB, AO =OB. In triangle COD, CO =OD.

So diagonals bisect each other.

Answer 2

Answer:

Quadrilaterals. Show that the diagonals of a square are equal and perpendicular to each other. ... ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that (i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D

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