Grade 1010 grade maths
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
1 Answer
Arun
In ΔABC, we have CD/DA = CE/EB as AB || DE
in ΔCDB, we have CF/FD = CE/EB as EF || BD
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
=> 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
Actually I have answered it before so I have directly put this
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