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Given -ABCD is a square and the diagonals AC and BD intersect at O.
To Prove- AO=OC and OB=OD and AC=BD
Proof- in ΔADC and ΔBCD
              ∠D=∠B=90 degrees
              CD =DC (common)
               AD=BC(sides of a square are equal)
               Therefore by SAS congruence rule ΔADC is congruent to ΔBcd
by cpct,
Therefore OC=OD(sides of an isoscles triangle)
since AC=BD,
Therefore in a square t5he diagonals are equal and they are bisected
Hence Proved

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Hello !
See a square  is the only quadrilateral which has all the sides equal and the all the angles between any two sides is 90°, and so it also has the two diagonals intersescting at 90°.
The prove of the intersection of the two diagonals :
        In a square the two  diagonals intersect they form four (4) equal small triangles within that square , taking one square the diagonal joining the vertex of the two sides of the square bisects the angle i.e the 90° is bisected which is equal to 45° and the other part is also 45° ,Taking two  angles i.e 45° and 45° and let the angle between the intersection of the diagonals be x° 
Hence, it is also known the diagonals intersect at the mid point of each other so the each  triangle is an isoceles 
Then , in any of the triangle ,
x° +45°+45°=180°
⇒ x°=180°- 90°
x° = 90°                                                    Ans.
Hence Proved,
I hope u are understood !
If any problem ask me in the message box.
Plz mark the answer as the brainliest if u are satisfied!!
Thank you.
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