Question

please answer my question ​

Answer 1

Hello again friend is it the same 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

Same answer for same question

Hope helps

Pls mark brainliest