B**I
发帖数: 601 | 1 In triangle ABC, Angle B is 120 degree, segment AF, BD and CE are angle
bisectors of angle A, B and C, and segment AF, BD and CE intercept segment
BC, AC and AB at point F, D and E. Prove Angle EDF is equal to 90 degree.
Thanks | g******a
发帖数: 69 | 2 Find a point G on EC such that BGC is similar to EFC.
Then EBC is similar to FGC, so that ∠EGF=60.
Thus EBGF is on the same circle, which implies
∠GBC=∠FEC=∠FBG=30. similarly, ∠ADF=30.
∠EFD=150-∠ADF-∠FEC=90.
【在 B**I 的大作中提到】
: In triangle ABC, Angle B is 120 degree, segment AF, BD and CE are angle
: bisectors of angle A, B and C, and segment AF, BD and CE intercept segment
: BC, AC and AB at point F, D and E. Prove Angle EDF is equal to 90 degree.
: Thanks
| B**I
发帖数: 601 | 3 Thanks a lot. But I do not understand. Could you make it clear?
and the purpose is to prove ∠EDF=90, maybe we drew different pictures.
【在 g******a 的大作中提到】
: Find a point G on EC such that BGC is similar to EFC.
: Then EBC is similar to FGC, so that ∠EGF=60.
: Thus EBGF is on the same circle, which implies
: ∠GBC=∠FEC=∠FBG=30. similarly, ∠ADF=30.
: ∠EFD=150-∠ADF-∠FEC=90.
| g******a
发帖数: 69 | 4 Oh, yes, I exchanged D and F.
【在 B**I 的大作中提到】
: Thanks a lot. But I do not understand. Could you make it clear?
: and the purpose is to prove ∠EDF=90, maybe we drew different pictures.
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