Question

Prove that the diagonals of a square are equal and bisect each other at 90°.​

Answer 1

Answer:

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.

HOPE THIS HELPS...............

Answer 2

$\bf\underline{\underline{\pink{Question:-}}}$

★ Prove that the diagonals of a square are equal and bisect each other at 90°.

$\bf\underline{\underline{\blue{Given:-}}}$

★ A square ABCD whose diagonals AC and BD intersect at O

$\bf\underline{\underline{\orange{To \: Prove:-}}}$

★ AC = BD

★ OA = OC and OB = OD

★ AC ⊥ BD

$\bf\underline{\underline{\green{Proof:-}}}$

★AC= BD

In △ABC and BAD, we have

AB = BA (common)

BC = AD (sides of square)

∠ABC = ∠BAD (each equal to 90°)

∴△ABC≅△BAD

And also, AC = BD (c.p.c.t)

★OA = OC and OZb = OD

Now, ABCD is a square and therefore a ||gm.

And so, OA = OC and OB = OD [∵diagonals of ||gm bisect each other]

★AC⊥BD

Now, in △AOB and AOD, we have

OB = OD [∵diagonals of ||gm bisect each other]

AB = AD (sides of square)

AO = AO (common)

∴△AOB ≅△AOD

∴ ∠AOB = ∠AOD

But, ∠AOB + ∠AOD = 180° (linear pair)

∠AOB = ∠AOD = 90°

Thus, AO⊥BD, i.e., AC⊥BD

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