Question -
Answer -
Let ABCD be a square and its diagonals AC and BD intersect eachother at O.
To show that,
AC = BD
AO = OC
and тИаAOB =90┬░
Proof,
In ╬ФABC and ╬ФBAD,
AB = BA (Common)
тИаABC = тИаBAD = 90┬░
BC = AD (Given)
╬ФABC тЙЕ ╬ФBAD[SAS congruency]
Thus,
AC = BD [CPCT]
diagonals are equal.
Now,
In ╬ФAOB and ╬ФCOD,
тИаBAO = тИаDCO (Alternate interior angles)
тИаAOB = тИаCOD (Vertically opposite)
AB = CD (Given)
, ╬ФAOB тЙЕ ╬ФCOD[AAS congruency]
Thus,
AO = CO [CPCT].
, Diagonal bisect each other.
Now,
In ╬ФAOB and ╬ФCOB,
OB = OB (Given)
AO = CO (diagonals are bisected)
AB = CB (Sides of the square)
, ╬ФAOB тЙЕ ╬ФCOB[SSS congruency]
also, тИаAOB = тИаCOB
тИаAOB+тИаCOB = 180┬░ (Linear pair)
Thus, тИаAOB = тИаCOB = 90┬░
,Diagonals bisect each other at right angles