In triangle ABC. AB is parallel to DE. BD is parallel to EF. Prove CD^2= CF*AC
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3See the diagram that was missing from the question.
In ΔABC, we have CD/DA = CE/EB as AB || DE
in ΔCDB, we have CF/FD = CE/EB as EF || BD
hence we get CD/DA = CF/FD
=> Reciprocals: DA/CD = FD /CF
=> Add 1 on both sides: AC /CD = DC/CF
=> DC^2 = CF * AC
In ΔABC, we have CD/DA = CE/EB as AB || DE
in ΔCDB, we have CF/FD = CE/EB as EF || BD
hence we get CD/DA = CF/FD
=> Reciprocals: DA/CD = FD /CF
=> Add 1 on both sides: AC /CD = DC/CF
=> DC^2 = CF * AC
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