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 Share with your friends Share 2 Tanveer Sofi answered this 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+AN2⇒PN2=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+PB2⇒AP2=AB2-PB2 ..........4Substituting the value of AP2 from equation 4 into equation 3, we getPB2=AB2-PB2-AN2+BN2⇒2PB2=AB2-AN2+BN2⇒2PB2=AB-ANAB+AN+BN2⇒2PB2=BNAB+AN+BN2 As, AB-AN=BN⇒2PB2=BNAB+AN+BN=BNAB+AB As, AN+BN=AB⇒2PB2=2BN×AB=2AB×BN⇒PB2=AB×BN. 1 View Full Answer
