Toolkit 41Dual Pythagorean TheoremCB2−CA2=EB2−EA2=FB2−FA2=DB2−DA2CB^2-CA^2=EB^2-EA^2=FB^2-FA^2=DB^2-DA^2CB2−CA2=EB2−EA2=FB2−FA2=DB2−DA2ProofBy applying the Pythagorean Theorem to triangles △BCD\triangle BCD△BCD and △ACD\triangle ACD△ACD,CB2=CD2+DB2,CB^2 = CD^2 + DB^2,CB2=CD2+DB2,CA2=CD2+DA2.CA^2 = CD^2 + DA^2.CA2=CD2+DA2.Therefore,CB2−CA2=DB2−DA2.CB^2 - CA^2 = DB^2 - DA^2.CB2−CA2=DB2−DA2.Similarly,EB2−EA2=DB2−DA2,EB^2 - EA^2 = DB^2 - DA^2,EB2−EA2=DB2−DA2,FB2−FA2=DB2−DA2.FB^2 - FA^2 = DB^2 - DA^2.FB2−FA2=DB2−DA2.Hence,CB2−CA2=EB2−EA2=FB2−FA2=DB2−DA2.□CB^2 - CA^2 = EB^2 - EA^2 = FB^2 - FA^2 = DB^2 - DA^2. \quad\squareCB2−CA2=EB2−EA2=FB2−FA2=DB2−DA2.□Related ProblemsCoreAMC 10A/12A 2025 (Problem 23/16)AMC 10A/12A 2022 (Problems 23/20)