AB is the diameter of a circle . From any point P on the circle the perpendicular drawn on AB meets AB at N. Prove that PB 2= AB * BN


As PB is perpendicular to AB, then in the right angled triangle PNB, we havePB2=PN2+BN2       ...........1In the right angled triangle PNA, we haveAP2=PN2+AN2PN2=AP2-AN2       ...........2From equation 1 and 2, we getPB2=AP2-AN2+BN2      ....................3Also, we know that the angle subtended by the diameter AB of the circle at any point P on the circle is 90. Therefore in right angled triangle APB, we haveAB2=AP2+PB2AP2=AB2-PB2          ..........4Substituting the value of AP2 from equation 4 into equation 3, we getPB2=AB2-PB2-AN2+BN22PB2=AB2-AN2+BN22PB2=AB-ANAB+AN+BN22PB2=BNAB+AN+BN2                                                 As, AB-AN=BN2PB2=BNAB+AN+BN=BNAB+AB                      As, AN+BN=AB2PB2=2BN×AB=2AB×BNPB2=AB×BN.

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