In the given figure, X and Y are centers of two intersecting equal circle. If $X Y = A B$ , prove that AXBY is a square.

Question

In the given figure, X and Y are centers of two intersecting equal circle. If XY=AB, prove that AXBY is a square.

vzletanje8z0 · Accepted Answer

Step 1It is given that X and Y are the centers of two intersecting equal circles by which we can say that the radius of both circles is equal.Two circles are equal if they have the same radius.Also, it is given that XY=ABTo Prove: AXBY is a SquareStep 2ProofSince the radius are equal Therefore we can say thatAX=AY=BY=XBSince all the sides are equal to a quadrilateral and also the diagonalsXY=ABAgain,In △AXB and △AX=AY (radius of same circle)BX=BY (radius of same circle)XY=AB (given)Therefore, By SSS (Side-Side-Side) congruence rule△AXB=∼△AYBBy CPCT∠AXB=∠AYBWe know the sum of the opposite angles of a quadrilateral is, Therefore, we have,∠AXB+∠AYB=180∘2∠AXB=180∘∠AXB=90∘Similarly,∠AXB=∠XBY=∠BYA=∠XAY=90∘Step 3All the sides of the quadrilateral AXBY are equal.Diagonals of the quadrilateral AXBY are equal.All angles of the quadrilateral AXBY are 90∘Hence AXBY is a squareHence Proved.

In the given figure, X and Y are centers of

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