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Solution: We have AE = AD … (i) and CE = BD … (ii) On adding (i) and (ii), we have AE + CE = AD + BD ⇒ AC = AB Now, in ∆AEB and ∆ADC, we have AE = AD [given] AB = AC [proved above] ∠A = ∠A [common] ∴ By SAS congruence axiom, we have ∆AEB = ∆ADC. microsoft. P = MCe^(kt) - PCe^(kt) P + PCe^(kt) = MCe^(kt) P[1 + Ce^(kt)] = MCe^(kt) P = MCe^(kt)/[1 + Ce^(kt)] Multiply the right hand side by [1/Ce^(kt)]/[1/Ce^(kt)] to get. Dans le sac de Marie, il y a un stylo-bille et une calculatrice. 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