↵In a quadrilateral ABCD , equal diagonals AC BD intersect at P, such that AP=PC BP=PD - Maths - Quadrilaterals

Question

↵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

Ashwini Kumar · Accepted Answer

Dear Student,

Please find below the solution to the asked query:
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In quadrilateral ABCD, AC and BD are equal diagonals and AP=PC & BP=OD∴AP=BP=CP=DP=a sayNow,∠BPC=90°⇒∠APD=90°      vertically opposite anglesHere,∠APD+∠CPD=180°      vertically opposite angles⇒90°+∠CPD=180°⇒∠CPD=90°Again∠APB+∠BPC=180°      vertically opposite angles⇒∠APB+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=DA∴ABCD 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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↵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