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v_Enhance, Dec 12, 2011, 6:56 AM
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This is trivialized with Ceva.
by
AwesomeToad, Dec 12, 2011, 6:53 PM
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Ceva
707
gives to us quite generously,
, or
, which implies
, which solves the problem. Therefore, it is most natural to use it.



707
by
AwesomeToad, Dec 12, 2011, 7:01 PM
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What he intended
Let G be the reflection of P over M, so BPCG is a parallelogram. There is a homothety centred around A sending ▲EPD to ▲BGC, so ▲EPD ~ ▲BGC ~ ▲BPC, so ∠PDE = ∠PBC.
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Yeah, that was it.
Alternatively
Alternatively
Just use parallel lines => similar triangles, after constructing
as above, to get
and the rest is obvious.


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