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## Answers

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,

AC=BD

∠BCD=∠ACD

Therefore OC=OD(sides of an isoscles triangle)

since AC=BD,

OA=OB

therefore,

OA=OB=OC=Od

Therefore in a square t5he diagonals are equal and they are bisected

Hence Proved

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 :

See,

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°

⇒

**Hence Proved,**

__x° = 90° Ans.__

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