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

It is given that ABCD is a rectangle

A = C


A = C


DAC = DCA (AC bisects A and C)



CD = DA (Sides opposite to equal angles are also equal)


However,


DA = BC and AB = CD (Opposite sides of a rectangle are equal)


AB = BC = CD = DA


ABCD is a rectangle and all of its sides are equal.


Hence, ABCD is a square


(ii) Let us join BD


In ΔBCD,


BC = CD (Sides of a square are equal to each other)


CDB = CBD (Angles opposite to equal sides are equal)


However,


CDB = ABD (Alternate interior angles for AB || CD)


CBD = ABD


BD bisects B


Also,


CBD = ADB (Alternate interior angles for BC || AD)


CDB = ABD


BD bisects D


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