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Hi there! Have questions about your homework? At Brainly, there are 60 million students who want to help each other learn. Questions are usually answered in less than 10 minutes. Try it for yourself by posting a question! :D

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, 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

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 : 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° ⇒ 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.