Question

please please help it would be very helpful​​

Answer 1

In ∆ ABC , 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

Hence, DC² = CF × AC

HOPE THIS HELPS UH.!

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