ABCD and PQRC are rectangles . Q is the midpoint of AC.Prove that DP=PC and PR=1÷2AC
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ΔACD similar to Δ QCP by AA test of similarity
∴AC/QC = DC/PC
∴(AC-QC)/QC = (DC-PC)/PC ... DIVIDENDO
∴AQ/QC =DP/PC ... A-Q-C , D-P-C
∴1 = DP/PC ... AQ=QC
∴DP =PC
Quadrilateral ABCD & PQRC are rectangles
∴QC = PR & QC = 1/2AC ...diagonals bisect each other ----(i)
∴PR = 1/2AC
∴AC/QC = DC/PC
∴(AC-QC)/QC = (DC-PC)/PC ... DIVIDENDO
∴AQ/QC =DP/PC ... A-Q-C , D-P-C
∴1 = DP/PC ... AQ=QC
∴DP =PC
Quadrilateral ABCD & PQRC are rectangles
∴QC = PR & QC = 1/2AC ...diagonals bisect each other ----(i)
∴PR = 1/2AC
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