↵In a quadrilateral ABCD , equal diagonals AC & BD intersect at P, such that AP=PC & BP=PD ,also angle BPC=90 degreee, then quadrilateral is exactly (a) a parallelogram (b) a square (c) a rhombus (d) a rectangle

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Please find below the solution to the asked query:

In quadrilateral ABCD, AC and BD are equal diagonals and AP=PC & BP=ODAP=BP=CP=DP=a sayNow,BPC=90°APD=90°      vertically opposite anglesHere,APD+CPD=180°      vertically opposite angles90°+CPD=180°CPD=90°AgainAPB+BPC=180°      vertically opposite anglesAPB+90°=180°APB=90°Therefore, APB=BPC=CPD=APD=90°In all four triangles; APB, BPC, CPD and APD,two sides equal to a and angles between these two sides are 90°hence, all four triangles are congruent.Ab=BC=CD=DAABCD is a rhobus.Now, a rhombus whose both diagonals are equal is square. ABCD is a square.Hence, b is correct option.
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