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In triangle abc DE parallel to BC and CD parallel to EF P.T (Ad)2 =AE ×AB

In  triangle abc DE parallel to BC and CD parallel to EF P.T (Ad)2 =AE ×AB

Grade:10

1 Answers

Arun
25750 Points
5 years ago
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
You can similarly prove relations like this.
Hope it helps
Actually I have answered it before so I have directly put this

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