Show that if the Diagonals of a Quadrilateral are equal and bisect eachother at right angles, then it is a square...?
Answers
Answer:
Refer diagram
We have a quadrilateral ABCD such that O is the mid-point of AC and BD. Also AC ⊥ BD.
Now, in ΔAOD and ΔAOB, we have
AO = AO
[Common]
OD = OB
[∵ O is the mid-point of BD]
∠AOD = ∠AOB
[Each = 90°]
∴ ΔAOD ≌ ∠AOB
[SAS criteria]
∴Their corresponding parts are equal.
⇒ AD = AB
...(1)
Similarly, we haveAB = BC...(2) BC = CD...(3) CD = DA...(4)
From (1), (2), (3) and (4) we have: AB = BC = CD = DA
∴Quadrilateral ABCD is having all sides equal.
In ΔAOD and ΔCOB, we have
AO = CO[Given] OD = OB[Given] ∠AOD = ∠COB[Vertically opposite angles] ∴ΔAOD ≌ ΔCOB
⇒Their corresponding pacts are equal.
⇒ ∠1 = ∠2
But, they form a pair of interior alternate angles.
∴AD || BC
Similarly, AB || DC
∴ ABCD is' a parallelogram.
∵ Parallelogram having all of its sides equal is a rhombus.
∴ ABCD is a rhombus.
Now, in ΔABC and ΔBAD, we have
AC = BD[Given] BC = AD[Proved] AB = BA[Common] ΔABC ≌ ΔBAD[SSS criteria]
Their corresponding angles are equal.
∠ABC = ∠BAD
Since, AD || BC and AB is a transversal.
∴∠ABC + ∠BAD = 180°[Interior opposite angles are supplementary]
i.e. The rhombus ABCD is having one angle equal to 90°.
Thus, ABCD is a square.
hope it helps :))
Answer:
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