# Show that if the diagonals of a quadrilateral are equal and bisect each other at a right angle,then it is a square.

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by monalisa

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by monalisa

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Let the quadrilateral be ABCD. Diagonals AC and BD iare equal and they ntersect (bisect) at right angles at O.

So AO = BO = OC = OD = AC/2 = BD/2

In ΔAOB and ΔBOC, angle AOB = 90 deg = BOC. AO = BO = CO

Hence both triangles are congruent. So AB = BC

Similarly in ΔBOC and Δ COD, they are congruent and BC = CD

Similarly in ΔCOD and Δ DOA, they are congruent and CD = DA

In Δ AOB, angle OAB = OBA as the sides OA and OB are equal.

angle OAB = angle OBA = (180 - AOB)/2 = 90/2 = 45 deg

Similarly in ΔBOC, ΔCOD and ΔDOA, we get

angle OBC = OCB = OCD = ODC = ODA = OAD = 45 deg.

Hence in quadrilateral ABCD, angle A = angle OAD + OAB = 45+45=90 deg.

Angle B = angle OBA + OBC = 45+45 = 90 deg.

Similarly other angles C = D = 90 deg.

Hence ABCD is a square.

So AO = BO = OC = OD = AC/2 = BD/2

In ΔAOB and ΔBOC, angle AOB = 90 deg = BOC. AO = BO = CO

Hence both triangles are congruent. So AB = BC

Similarly in ΔBOC and Δ COD, they are congruent and BC = CD

Similarly in ΔCOD and Δ DOA, they are congruent and CD = DA

In Δ AOB, angle OAB = OBA as the sides OA and OB are equal.

angle OAB = angle OBA = (180 - AOB)/2 = 90/2 = 45 deg

Similarly in ΔBOC, ΔCOD and ΔDOA, we get

angle OBC = OCB = OCD = ODC = ODA = OAD = 45 deg.

Hence in quadrilateral ABCD, angle A = angle OAD + OAB = 45+45=90 deg.

Angle B = angle OBA + OBC = 45+45 = 90 deg.

Similarly other angles C = D = 90 deg.

Hence ABCD is a square.